{"version":1,"lectureId":"01M1G55RAYY6BFTNC2PZYEQ2CX","attempt":0,"publication":{"slug":"why-a-a-is-secretly-a-theorem","title":"Why −(−a) = a Is Secretly a Theorem","subject":"mathematics","summary":"Everybody knows that two minus signs make a plus, but in the first chapter of a real analysis course it is a theorem, and its proof is one motion. This lecture reads the minus sign as a job title rather than a sign: minus a is whichever number cancels a, so the double minus is the canceler of the canceler. The inverse axiom promises that a canceler exists; it never promises there is only one. We build a four element addition table where identity, inverses and commutativity all hold, associativity fails, and the double minus genuinely lands on the wrong number. Then we slide one bracket along a row of three symbols to prove that cancelers are unique, apply the same slide to a plus its canceler plus that canceler's canceler, and finish by marking out what this ground floor result does not yet give you: nothing about products, and nothing about the word positive.","metaDescription":"Two minus signs make a plus is a theorem: existence is free from the inverse axiom, uniqueness costs one slide of a bracket.","transcript":"Everybody knows this one. Turn around, then turn around again, and you are facing the way you started. Flip a coin twice and it shows the same face. Cancel somebody's debt, and you have handed them money. Two minus signs make a plus. So why did one short question about it turn into a twelve hour argument, on a forum full of people learning mathematics? An argument that ended only when one of the two people in it wrote, you are right, my apologies. The two sides went like this. One of them said the fact holds in any system where you can add and cancel, so there is nothing to prove. The other said that the proof on offer assumed the very thing it was trying to prove. You cannot get a theorem by writing a symbol down in a suggestive way. They were both pointing at something real. There is a small, genuine theorem hiding inside this obvious fact, and the argument was about where it lives. To find it we have to work only from the rules. The person who asked was reading the first chapter of Spivak's Calculus, which opens by writing down the rules that numbers obey. From here on we are allowed exactly what the rules say and nothing else. No obviously. No number line. Rule one: grouping does not matter. If you are adding three things, you may bracket the first pair or the last pair, and you get the same answer. Rule two: there is a number called zero, and adding it does nothing at all. And rule three, the one the whole argument is about. For every number a, there is a number, written minus a, that cancels it. a plus minus a is zero, and minus a plus a is zero as well. Now read that third rule again, slowly, in words. Four of those words are going to carry this entire video. There is a number. It says there is a number. It never says there is only one. And notice one last thing before we go on. The fourth rule on that list, that a plus b equals b plus a, we are never going to use. Three rules and a claim. Before we argue about the claim, we had better be sure what its symbols mean, and the minus sign is where people slip. In this rulebook the minus sign is not a property of a number. It is not negativeness. It is a job title. Minus a means whichever number holds the job of cancelling a. Rule three hands us that number and says what the job is: a plus minus a is zero. Let me draw it. Here is a, here is its canceler, and I will join two numbers with a line whenever they add to zero. If a is five, the canceler is minus five. If a is minus three, the canceler is three, so a symbol with a minus in front can perfectly well be a positive number. And zero cancels itself. Now read the thing we are trying to prove from the inside out. Minus a is the canceler of a. So minus minus a is the canceler of the canceler of a. I am keeping that one grey, with a question mark on it, because at this point in the story we do not know which number it is. Here is where the thread got tangled, and it is a good tangle. Does a cancel minus a? Yes, at once. Rule three says minus a plus a is zero, and that is exactly the job. So a does it, for free. So are we done? Not quite, and the difference is a single word. What we have shown is that a is a canceler of minus a. What the symbol names is the canceler of minus a. Look at rule three once more. It promised that a canceler exists. It never promised there is only one. If minus a had two different cancelers, the symbol would be ambiguous. It might be a. It might be the grey one. Two minuses making a plus would be a coin toss. So here is the theorem underneath the claim. Every number has exactly one canceler. Once you have that, minus minus a equals a is only that theorem, read out at the number minus a. And you might think exactly one canceler is obviously true. It is not. To see that exactly one canceler is not obvious, let me build a number system where it is false. Four numbers: zero, a, b and c. Its addition is given by a table, the same kind you learned your sums from, only smaller. Check rule two. The zero row and the zero column hand everything straight back, unchanged. Zero does nothing here. Check rule three. Every number here has a canceler. a and b add to zero. c and b add to zero. And zero cancels itself. The table is even symmetric, so this addition commutes too. Now look at the row for b. It has two zeros in it. b is cancelled by a, and b is also cancelled by c. In this world, b has two cancelers. So compute the canceler of the canceler of a. The row for a has exactly one zero, in the b column, so minus a is b. Now the row for b has two zeros, so the canceler of b is a, or it is c. Take c. In this little world, minus minus a is c, and c is not a. Which rule did we break? Not rule two. Not rule three. And commutativity is fine as well. Look at rule one. Add a and b first, then add c: a plus b is zero, and zero plus c is c. Now bracket the other pair: b plus c is zero, and a plus zero is a. And look at the two values we just got. The first was c. The second was a. Those are exactly the two numbers that were fighting over the title canceler of b. That is not a coincidence. It is our theorem, seen from the wrong side. So what does our number system have that the toy one does not? One thing. The bracket can slide. Here is what that buys us. Suppose two numbers, a and c, both cancel the same number b. So a plus b is zero, and b plus c is zero. I am assuming nothing else about them. Must a and c be the same number? Write all three of them in a row: a plus b plus c. Rule one says this row has one value, no matter how I bracket it. So let us bracket it two different ways and compare. Bracket the left pair first. a and b cancel, so that pair is zero. Zero plus c is c. The whole row is worth c. Now the one motion to remember from this whole video. Watch the bracket leave the left pair, and land on the right pair instead. Now the same thing happens at the other end. b and c cancel, so that pair is zero. a plus zero is a. The row is worth a. Same row. One value. So a equals c, and that is the proof. Any two cancelers of the same number are equal. Cancelers are unique. There is nothing clever in that argument. a and c only had to be in the same room as b, and rule one is what puts them there. In the toy world the bracket cannot slide, so a and c never have to meet, and nothing makes them equal. That is the only difference between the two worlds. Let us write that down as a theorem, because we are about to use it twice. If a cancels b, and c also cancels b, then a and c are the same number. Rule three says a canceler exists. Rule one says there is only ever one. Before we cash it in, let us fix what the symbol is allowed to mean. Minus minus a is a canceler of minus a, any one of them. We are not assuming there is only one of them, because that is the thing we are proving. Now rule three gives us two promises. First: a has a canceler, and the two of them add to zero. Second: minus a has a canceler too, and minus minus a is the name of one. Write all three in a row, exactly as before: a, then the canceler of a, then the canceler of that. Rule one says this row has one value however I bracket it. Bracket the left pair. a and its canceler make zero, and zero plus the last slot leaves the last slot standing. So the row is worth minus minus a. Now slide it. The bracket leaves the left pair and lands on the right pair instead. And now the other two cancel, by the promise we just wrote down. a plus zero is a. So the row is worth a. Same row, one value. So minus minus a equals a. That is the theorem we came for, and it is now proved. Written as a single chain, the whole proof is five lines. Zero does nothing, so the thing on the left is zero plus itself. That is rule two. That zero is a plus the canceler of a: rule three. And now the only interesting line in the proof. Slide the bracket. Rule one. The middle and the right cancel, by rule three again, leaving a plus zero. And a plus zero is a, by rule two. Done. Now notice what never happened in that chain. We never swapped two terms. The fourth rule, that a plus b equals b plus a, was never used. So this is not really a fact about numbers. It holds anywhere there is a zero, an undo for every move, and a bracket that slides. Rotating a cube. Shuffling a deck. The undo of the undo is the move you started with. So back to the argument. Who was right? Both of them, describing one theorem from opposite ends. The person who asked had the shape of it exactly right. a cancels minus a. Cancelers are unique. Therefore a is the canceler of minus a, and minus minus a is only its name. The objection was right too: you cannot get that middle line by notation. Which is exactly why the small proof earns its keep. It proves the middle line on the spot, with one slide of the bracket. The objector saw that, and said so. That is a good thread. One more question from the thread, and it is a good one. Does everything we have just done prove that a negative times a negative is a positive? No. And seeing why is worth a minute. Think of the rules as a building. Everything today happened on the ground floor: addition, zero, cancelers, and the bracket that slides. On this floor there is no multiplication at all, and no such thing as a positive or a negative number. Minus a is only the canceler of a, and it might be five, or minus five, or zero. To even say negative times negative you need the next floor up: multiplication, and the rule that ties it to addition. Watch what happens there. Add a to minus one times a, and rewrite the first a as one times a. Pull the a out. One plus minus one is zero, and zero times anything is zero. So minus one times a cancels a. And because cancelers are unique, minus one times a is minus a. That is our engine, borrowed. From there, minus a times minus b works out to a times b, and that step uses today's theorem. But you still have not said the word positive. For that you need a third floor, the rules for which numbers count as positive, and only there does the sentence mean anything. Today's theorem is one brick in that wall. A load bearing one. But it is not the wall. So here is the picture to keep. Draw a line between two numbers whenever they add to zero. Rule three says every number gets at least one such line. Rule one says no number gets two. So the lines make a perfect pairing, and zero is its own partner. Minus minus a equals a says exactly this: go to your partner, then go to your partner's partner, and you are home. Turn around twice: true. Why it is true: one slide of a bracket.","watch":{"version":1,"scenes":[{"title":"Two Minuses, and an Argument","start":0,"end":130.90145833333332,"objects":{"chapter":"a Tex [text] that says \"Spivak, Chapter 1: Basic Properties of Numbers\"","claim":"a Math [text] that says \"$-(-a) = a$\"","head_open":"a Heading that says \"Everybody Knows This One\"","head_rules":"a Heading that says \"The Rulebook\"","head_thread":"a Heading that says \"A Twelve Hour Argument\"","intuitions":"a Block [text] that says \"Turn around, then turn around again. Flip a coin twice. Cancel a debt, and you are given money.\"","label_no":"a Tex [text] that says \"Or did it need a proof?\"","label_yes":"a Tex [text] that says \"Was it obvious?\"","p1":"a Math [text] that says \"$upright(\"P1\") quad a + (b + c) = (a + b) + c$\"","p2":"a Math [text] that says \"$upright(\"P2\") quad a + 0 = 0 + a = a$\"","p3":"a Math [text] that says \"$upright(\"P3\") quad a + (-a) = (-a) + a = 0$\"","p3_words":"a Panel that says \"For every number $a$ there is a number $-a$ such that $a + (-a) = (-a) + a = 0$.\"","p4_aside":"a Panel that says \"The list has a fourth rule, $a + b = b + a$. We will not need it.\"","voice_no":"a Panel that says \"Your proof assumes its own conclusion. Writing a symbol down in a suggestive way is not an argument.\"","voice_yes":"a Panel that says \"This holds in every system where you can add and cancel. There is nothing here to prove.\""},"beats":[{"start":0,"say":"Everybody knows this one. Turn around, then turn around again, and you are facing the way you started. Flip a coin twice and it shows the same face. Cancel somebody's debt, and you have handed them money. Two minus signs make a plus.","live":[],"does":[[0,"head_open is shown on the screen, written out."],[0,"claim is shown on the screen, written out."],[1.741,"intuitions is shown on the screen, written out."],[5.422,"intuitions (the \"Turn around\" part) is emphasized."],[7.21,"intuitions (the \"Flip a coin\" part) is emphasized."],[7.21,"intuitions (the \"Turn around\" part) is no longer emphasized."],[10.321,"intuitions (the \"Cancel a debt\" part) is emphasized."],[10.321,"intuitions (the \"Flip a coin\" part) is no longer emphasized."],[14.594,"intuitions (the \"Cancel a debt\" part) is no longer emphasized."]]},{"start":15.193999999999999,"say":"So why did one short question about it turn into a twelve hour argument, on a forum full of people learning mathematics? An argument that ended only when one of the two people in it wrote, you are right, my apologies.","live":["claim","intuitions","head_open"],"does":[[16.702999999999996,"claim is indicated — a transient flash."],[28.011499999999998,"claim is hidden from the screen — left the board."],[28.011499999999998,"head_open is hidden from the screen — left the board."],[28.011499999999998,"intuitions is hidden from the screen — left the board."]]},{"start":29.2115,"say":"The two sides went like this. One of them said the fact holds in any system where you can add and cancel, so there is nothing to prove.","live":[],"does":[[29.2115,"head_thread is shown on the screen, written out."],[31.881,"label_yes is shown on the screen, written out."],[32.415000000000006,"voice_yes is shown on the screen, written out."]]},{"start":38.008,"say":"The other said that the proof on offer assumed the very thing it was trying to prove. You cannot get a theorem by writing a symbol down in a suggestive way.","live":["label_yes","voice_yes","head_thread"],"does":[[38.008,"label_no is shown on the screen, written out."],[40.376000000000005,"voice_no is shown on the screen, written out."]]},{"start":47.1815,"say":"They were both pointing at something real. There is a small, genuine theorem hiding inside this obvious fact, and the argument was about where it lives. To find it we have to work only from the rules.","live":["label_yes","voice_yes","label_no","voice_no","head_thread"],"does":[[47.686,"voice_yes (the \"add and cancel\" part) is indicated — a transient flash."],[48.836,"voice_no (the \"assumes its own conclusion\" part) is indicated — a transient flash."],[59.865,"head_thread is hidden from the screen — left the board."],[59.865,"label_no is hidden from the screen — left the board."],[59.865,"label_yes is hidden from the screen — left the board."],[59.865,"voice_no is hidden from the screen — left the board."],[59.865,"voice_yes is hidden from the screen — left the board."]]},{"start":61.065,"say":"The person who asked was reading the first chapter of Spivak's Calculus, which opens by writing down the rules that numbers obey. From here on we are allowed exactly what the rules say and nothing else. No obviously. No number line.","live":[],"does":[[61.065,"head_rules is shown on the screen, written out."],[63.306,"chapter is shown on the screen, written out."]]},{"start":76.502,"say":"Rule one: grouping does not matter. If you are adding three things, you may bracket the first pair or the last pair, and you get the same answer.","live":["chapter","head_rules"],"does":[[77.326,"p1 is shown on the screen, written out."]]},{"start":86.564,"say":"Rule two: there is a number called zero, and adding it does nothing at all.","live":["chapter","p1","head_rules"],"does":[[87.261,"p2 is shown on the screen, written out."]]},{"start":92.6325,"say":"And rule three, the one the whole argument is about. For every number a, there is a number, written minus a, that cancels it. a plus minus a is zero, and minus a plus a is zero as well.","live":["chapter","p1","p2","head_rules"],"does":[[93.53800000000001,"p3 is shown on the screen, written out."],[99.73800000000001,"p3 (the \"(-a)\" part) is indicated — a transient flash."]]},{"start":107.4195,"say":"Now read that third rule again, slowly, in words. Four of those words are going to carry this entire video. There is a number.","live":["chapter","p1","p2","p3","head_rules"],"does":[[110.635,"p3_words is shown on the screen, written out."],[113.31700000000001,"p3_words (the \"there is a number\" part) is emphasized."]]},{"start":116.645,"say":"It says there is a number. It never says there is only one. And notice one last thing before we go on. 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It is a name for whichever number cancels $a$.\"","jobs":"a Table [text] that says \"Name Its job $-a$ cancels $a$ $-(-a)$ cancels $-a$\" (rows=(('Name', 'Its job'), ('$-a$', 'cancels $a$'), ('$-(-a)$', 'can…, header=True)","node_a":"a Point [blue] labelled \"a\" drawn in web (location=(1.5, 3.0))","node_na":"a Point [yellow] labelled \"-a\" drawn in web (location=(4.9, 3.0))","node_q":"a Point [gray] labelled \"?\" drawn in web (location=(4.9, 1.0))","target":"a Math [gray] that says \"$-(-a) = a$\"","under":"a Panel that says \"Every number has exactly one canceler.\"","web":"a Figure (x_range=(0.0, 6.4), y_range=(0.0, 4.2), aspect=(5.0, 3.4))"},"beats":[{"start":130.90145833333332,"say":"Three rules and a claim. Before we argue about the claim, we had better be sure what its symbols mean, and the minus sign is where people slip.","live":[],"does":[[130.90145833333332,"head_job is shown on the screen, written out."]]},{"start":140.1159583333333,"say":"In this rulebook the minus sign is not a property of a number. It is not negativeness. It is a job title. Minus a means whichever number holds the job of cancelling a.","live":["head_job"],"does":[[146.43145833333332,"job is shown on the screen, written out."]]},{"start":152.42995833333333,"say":"Rule three hands us that number and says what the job is: a plus minus a is zero. Let me draw it. Here is a, here is its canceler, and I will join two numbers with a line whenever they add to zero.","live":["job","head_job"],"does":[[158.74545833333332,"cancels is shown on the screen, written out."],[160.35945833333332,"web is shown on the screen, written out."],[161.64845833333334,"node_a is shown on the screen, written out."],[163.2504583333333,"node_na is shown on the screen, written out."],[165.88545833333333,"edge is shown on the screen, written out."]]},{"start":168.78445833333333,"say":"If a is five, the canceler is minus five. If a is minus three, the canceler is three, so a symbol with a minus in front can perfectly well be a positive number. And zero cancels itself.","live":["job","cancels","web","head_job","node_a","node_na","edge"],"does":[[170.63045833333334,"node_na is indicated — a transient flash."],[183.48245833333334,"web moves to a new place on the board."],[183.48245833333334,"cancels is hidden from the screen — left the board."],[183.48245833333334,"head_job is hidden from the screen — left the board."],[183.48245833333334,"job is hidden from the screen — left the board."]]},{"start":184.68245833333333,"say":"Now read the thing we are trying to prove from the inside out. Minus a is the canceler of a. So minus minus a is the canceler of the canceler of a.","live":["web","node_a","node_na","edge"],"does":[[184.68245833333333,"head_parse is shown on the screen, written out."],[184.68245833333333,"jobs is shown on the screen, written out."],[189.16345833333332,"jobs is shown on the screen, written out."],[192.61145833333333,"jobs is shown on the screen, written out."],[194.09845833333333,"node_q is shown on the screen, written out."],[194.51647242951907,"edge_q is shown on the screen, written out."]]},{"start":197.19395833333334,"say":"I am keeping that one grey, with a question mark on it, because at this point in the story we do not know which number it is.","live":["web","node_a","node_na","edge","head_parse","node_q","edge_q"],"does":[[200.1314583333333,"node_q is indicated — a transient flash."],[205.45995833333333,"head_parse is hidden from the screen — left the board."],[205.45995833333333,"jobs is hidden from the screen — left the board."]]},{"start":206.65995833333332,"say":"Here is where the thread got tangled, and it is a good tangle. Does a cancel minus a? Yes, at once. Rule three says minus a plus a is zero, and that is exactly the job. So a does it, for free.","live":["web","node_a","node_na","edge","node_q","edge_q"],"does":[[206.65995833333332,"head_gap is shown on the screen, written out."],[214.96145833333333,"free is shown on the screen, written out."],[221.54445833333332,"edge is indicated — a transient flash."]]},{"start":222.86445833333332,"say":"So are we done? Not quite, and the difference is a single word. What we have shown is that a is a canceler of minus a. What the symbol names is the canceler of minus a.","live":["web","node_a","node_na","edge","node_q","edge_q","free","head_gap"],"does":[[227.10245833333335,"gap_q is shown on the screen, written out."]]},{"start":235.5849583333333,"say":"Look at rule three once more. It promised that a canceler exists. It never promised there is only one. If minus a had two different cancelers, the symbol would be ambiguous. It might be a. It might be the grey one. 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Every number has exactly one canceler. Once you have that, minus minus a equals a is only that theorem, read out at the number minus a. And you might think exactly one canceler is obviously true. 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Its plus sits in a circle, so nobody mistakes it for ordinary addition.\"","verdict":"a Math [text] that says \"$c eq.not a$\""},"beats":[{"start":273.97304166666663,"say":"To see that exactly one canceler is not obvious, let me build a number system where it is false. Four numbers: zero, a, b and c. Its addition is given by a table, the same kind you learned your sums from, only smaller.","live":[],"does":[[273.97304166666663,"toy_q is shown on the screen, written out."],[285.64104166666664,"table is shown on the screen, written out."],[286.37352270382706,"table is shown on the screen, written out."],[286.6664936282233,"table is shown on the screen, written out."],[286.98596021086075,"table is shown on the screen, written out."],[287.2145458418675,"table is shown on the screen, written out."]]},{"start":290.0840416666666,"say":"Check rule two. The zero row and the zero column hand everything straight back, unchanged. Zero does nothing here.","live":["toy_q"],"does":[[290.43204166666663,"f_zero is shown on the screen, written out."],[292.65004166666665,"table (the \"row=2\" part) is emphasized."],[293.6950416666666,"table (the \"column=2\" part) is emphasized."],[293.6950416666666,"table (the \"row=2\" part) is no longer emphasized."],[297.89804166666664,"table (the \"column=2\" part) is no longer emphasized."]]},{"start":299.5720416666666,"say":"Check rule three. Every number here has a canceler. a and b add to zero. c and b add to zero. And zero cancels itself. The table is even symmetric, so this addition commutes too.","live":["f_zero","toy_q"],"does":[[303.79204166666665,"f_ab is shown on the screen, written out."],[305.95104166666664,"f_cb is shown on the screen, written out."]]},{"start":314.48154166666666,"say":"Now look at the row for b. It has two zeros in it. b is cancelled by a, and b is also cancelled by c. In this world, b has two cancelers.","live":["f_zero","f_ab","f_cb","toy_q"],"does":[[315.48004166666664,"table (the \"row=4\" part) is emphasized."],[318.9980416666666,"f_ab is indicated — a transient flash."],[320.4140416666666,"f_cb is indicated — a transient flash."],[325.2090416666666,"table (the \"row=4\" part) is no longer emphasized."]]},{"start":325.80904166666664,"say":"So compute the canceler of the canceler of a. The row for a has exactly one zero, in the b column, so minus a is b. Now the row for b has two zeros, so the canceler of b is a, or it is c. Take c. In this little world, minus minus a is c, and c is not a.","live":null,"does":[[329.7090416666666,"table (the \"row=3\" part) is emphasized."],[332.9020416666666,"f_na is shown on the screen, written out."],[334.56204166666663,"table (the \"row=3\" part) is no longer emphasized."],[334.56204166666663,"table (the \"row=4\" part) is emphasized."],[335.5720416666666,"f_nb is shown on the screen, written out."],[341.22704166666665,"toy_answer is shown on the screen, written out."],[344.3030416666667,"table (the \"row=4\" part) is no longer emphasized."],[345.15054166666664,"f_ab is hidden from the screen — left the board."],[345.15054166666664,"f_cb is hidden from the screen — left the board."],[345.15054166666664,"f_na is hidden from the screen — left the board."],[345.15054166666664,"f_nb is hidden from the screen — left the board."],[345.15054166666664,"f_zero is hidden from the screen — left the board."],[345.15054166666664,"table is hidden from the screen — left the board."],[345.15054166666664,"toy_answer is hidden from the screen — left the board."],[345.15054166666664,"toy_q is hidden from the screen — left the board."]]},{"start":346.35054166666663,"say":"Which rule did we break? Not rule two. Not rule three. And commutativity is fine as well. Look at rule one. Add a and b first, then add c: a plus b is zero, and zero plus c is c. Now bracket the other pair: b plus c is zero, and a plus zero is a.","live":[],"does":[[346.35054166666663,"head_break is shown on the screen, written out."],[346.35054166666663,"checklist is shown on the screen, written out."],[348.90504166666665,"checklist (the \"P2 holds\" part) is emphasized."],[350.3100416666666,"checklist (the \"P2 holds\" part) is no longer emphasized."],[350.3100416666666,"checklist (the \"P3 holds\" part) is emphasized."],[351.6570416666666,"checklist (the \"Addition even commutes\" part) is emphasized."],[351.6570416666666,"checklist (the \"P3 holds\" part) is no longer emphasized."],[356.20804166666665,"assoc is shown on the screen, written out."],[363.15104166666663,"assoc is shown on the screen, written out."],[364.85704166666665,"checklist (the \"Addition even commutes\" part) is no longer emphasized."],[364.85704166666665,"checklist (the \"P1 fails\" part) is emphasized."]]},{"start":367.8025416666666,"say":"And look at the two values we just got. The first was c. The second was a. Those are exactly the two numbers that were fighting over the title canceler of b. That is not a coincidence. It is our theorem, seen from the wrong side.","live":["checklist","head_break"],"does":[[368.73104166666667,"verdict is shown on the screen, written out."],[370.28704166666665,"assoc is indicated — a transient flash."],[371.71504166666665,"assoc is indicated — a transient flash."],[374.86104166666667,"verdict is indicated — a transient flash."],[377.8100416666666,"checklist (the \"P1 fails\" part) is no longer emphasized."],[381.5022291666666,"assoc is hidden from the screen — left the board."],[381.5022291666666,"checklist is hidden from the screen — left the board."],[381.5022291666666,"head_break is hidden from the screen — left the board."],[381.5022291666666,"verdict is hidden from the screen — left the board."]]}]},{"title":"The Bracket Slides","start":382.5438958333333,"end":502.8800625,"objects":{"ask":"a Tex [text] that says \"Both $a$ and $c$ cancel $b$. Must they be the same number?\"","head_slide":"a Heading that says \"The Bracket Slides\"","head_unique":"a Heading that says \"Cancelers Are Unique\"","readings":"a Derivation [text] that says \"$(a + b) + c &= 0 + c = c \\ a + (b + c) &= a + 0 = a$\"","row":"a Math [text] that says \"$a + b + c$\"","same":"a Math [text] that says \"$a = c$\"","setup":"a Math [text] that says \"$a + b = 0, quad b + c = 0$\"","uniq":"a Panel that says \"A number has at most one canceler. With P3, it has exactly one.\"","uniq_claim":"a Math [text] that says \"$a + b = 0 thin upright(\"and\") thin b + c = 0 quad arrow.r quad a = c$\""},"beats":[{"start":382.5438958333333,"say":"So what does our number system have that the toy one does not? One thing. The bracket can slide. Here is what that buys us.","live":[],"does":[[382.5438958333333,"head_slide is shown on the screen, written out."],[390.4618958333333,"ask is shown on the screen, written out."]]},{"start":392.0773958333333,"say":"Suppose two numbers, a and c, both cancel the same number b. So a plus b is zero, and b plus c is zero. I am assuming nothing else about them. Must a and c be the same number?","live":["ask","head_slide"],"does":[[398.77089583333327,"setup is shown on the screen, written out."],[404.7378958333333,"ask is indicated — a transient flash."]]},{"start":407.5323958333333,"say":"Write all three of them in a row: a plus b plus c. Rule one says this row has one value, no matter how I bracket it. So let us bracket it two different ways and compare.","live":["ask","setup","head_slide"],"does":[[408.8788958333333,"row is shown on the screen, written out."]]},{"start":420.3998958333333,"say":"Bracket the left pair first. a and b cancel, so that pair is zero. Zero plus c is c. The whole row is worth c.","live":["ask","setup","row","head_slide"],"does":[[421.3638958333333,"row becomes \"$(a + b) + c$\"."],[427.2728958333333,"readings is shown on the screen, written out."]]},{"start":432.5543958333333,"say":"Now the one motion to remember from this whole video. Watch the bracket leave the left pair, and land on the right pair instead.","live":null,"does":[[437.0478958333333,"row becomes \"$a + (b + c)$\"."]]},{"start":441.5583958333333,"say":"Now the same thing happens at the other end. b and c cancel, so that pair is zero. a plus zero is a. The row is worth a.","live":null,"does":[[445.5868958333333,"readings is shown on the screen, written out."]]},{"start":452.09839583333326,"say":"Same row. One value. So a equals c, and that is the proof. Any two cancelers of the same number are equal. Cancelers are unique.","live":null,"does":[[454.9018958333333,"same is shown on the screen, written out."],[456.5968958333333,"A box is drawn around same."]]},{"start":463.8753958333333,"say":"There is nothing clever in that argument. a and c only had to be in the same room as b, and rule one is what puts them there. In the toy world the bracket cannot slide, so a and c never have to meet, and nothing makes them equal. That is the only difference between the two worlds.","live":["ask","setup","row","same","head_slide"],"does":[[469.0998958333333,"row is indicated — a transient flash."],[477.4358958333333,"readings is indicated — a transient flash."],[482.9388958333333,"ask is hidden from the screen — left the board."],[482.9388958333333,"head_slide is hidden from the screen — left the board."],[482.9388958333333,"readings is hidden from the screen — left the board."],[482.9388958333333,"row is hidden from the screen — left the board."],[482.9388958333333,"same is hidden from the screen — left the board."],[482.9388958333333,"setup is hidden from the screen — left the board."]]},{"start":484.1388958333333,"say":"Let us write that down as a theorem, because we are about to use it twice. If a cancels b, and c also cancels b, then a and c are the same number. Rule three says a canceler exists. Rule one says there is only ever one.","live":[],"does":[[484.1388958333333,"head_unique is shown on the screen, written out."],[489.06189583333327,"uniq_claim is shown on the screen, written out."],[495.3658958333333,"uniq is shown on the screen, written out."],[500.35789583333326,"uniq (the \"exactly one\" part) is emphasized."],[501.58839583333327,"uniq (the \"exactly one\" part) is no longer emphasized."],[501.83839583333327,"head_unique is hidden from the screen — left the board."],[501.83839583333327,"uniq is hidden from the screen — left the board."],[501.83839583333327,"uniq_claim is hidden from the screen — left the board."]]}]},{"title":"Cashing It In","start":502.8800625,"end":677.8892708333333,"objects":{"answer":"a Math [text] that says \"$-(-a) = a$\"","chain":"a Derivation [text] that says \"$-(-a) &= 0 + (-(-a)) quad upright(\"(P2)\") \\ &= (a + (-a)) + (-(-a)) quad upright(\"(P3)\") \\ &= a + ((-a) + (-(-a))) quad upright(\"(P1)\") \\ &= a + 0 quad upright(\"(P3)\") \\ &= a quad upright(\"(P2)\")$\"","fact_a":"a Math [text] that says \"$a + (-a) = 0$\"","fact_na":"a Math [text] that says \"$(-a) + (-(-a)) = 0$\"","head_cash":"a Heading that says \"Cash It In\"","head_chain":"a Heading that says \"The Whole Proof, in Five Steps\"","head_who":"a Heading that says \"So Who Was Right?\"","honest":"a Panel that says \"From here on $-(-a)$ means only this: a canceler of $-a$, any one of them. We are not assuming it is unique. That is what we are proving.\"","label_claim":"a Tex [text] that says \"What the claim contains\"","label_thread":"a Tex [text] that says \"How the thread ended\"","p4":"a Math [text] that says \"$a + b = b + a$\"","p4_note":"a Tex [text] that says \"Never used.\"","readings2":"a Derivation [text] that says \"$(a + (-a)) + (-(-a)) &= 0 + (-(-a)) = -(-a) \\ a + ((-a) + (-(-a))) &= a + 0 = a$\"","recipe":"a Block [text] that says \"$a$ cancels $-a$. Free: that is P3. Cancelers are unique. One slide of the bracket. So $a$ is the canceler of $-a$, and $-(-a)$ is its name.\"","row2":"a Math [text] that says \"$a + (-a) + (-(-a))$\"","voice_end":"a Panel that says \"You are right. That is exactly what it shows. My apologies.\"","voice_obj":"a Panel that says \"Proof by notation is not valid.\""},"beats":[{"start":502.8800625,"say":"Before we cash it in, let us fix what the symbol is allowed to mean. Minus minus a is a canceler of minus a, any one of them. We are not assuming there is only one of them, because that is the thing we are proving.","live":[],"does":[[502.8800625,"head_cash is shown on the screen, written out."],[505.7940625,"honest is shown on the screen, written out."],[513.3170625,"honest (the \"any one of\" part) is emphasized."],[516.8235625,"honest (the \"any one of\" part) is no longer emphasized."]]},{"start":517.4235625,"say":"Now rule three gives us two promises. First: a has a canceler, and the two of them add to zero. Second: minus a has a canceler too, and minus minus a is the name of one.","live":["honest","head_cash"],"does":[[520.6860625,"fact_a is shown on the screen, written out."],[525.2140625,"fact_na is shown on the screen, written out."]]},{"start":531.3515625,"say":"Write all three in a row, exactly as before: a, then the canceler of a, then the canceler of that. Rule one says this row has one value however I bracket it.","live":["honest","fact_a","fact_na","head_cash"],"does":[[532.7330625,"row2 is shown on the screen, written out."]]},{"start":543.6385625,"say":"Bracket the left pair. a and its canceler make zero, and zero plus the last slot leaves the last slot standing. So the row is worth minus minus a.","live":["honest","fact_a","fact_na","row2","head_cash"],"does":[[544.3350625,"row2 becomes \"$(a + (-a)) + (-(-a))$\"."],[548.1660625,"readings2 is shown on the screen, written out."]]},{"start":555.1195625,"say":"Now slide it. The bracket leaves the left pair and lands on the right pair instead.","live":null,"does":[[557.3490625000001,"row2 becomes \"$a + ((-a) + (-(-a)))$\"."]]},{"start":561.2020625,"say":"And now the other two cancel, by the promise we just wrote down. a plus zero is a. So the row is worth a.","live":null,"does":[[562.5020625,"readings2 is shown on the screen, written out."]]},{"start":569.9365625,"say":"Same row, one value. So minus minus a equals a. That is the theorem we came for, and it is now proved.","live":null,"does":[[573.6400625,"answer is shown on the screen, written out."],[575.4860625,"A box is drawn around answer."],[578.2255625,"answer is hidden from the screen — left the board."],[578.2255625,"fact_a is hidden from the screen — left the board."],[578.2255625,"fact_na is hidden from the screen — left the board."],[578.2255625,"head_cash is hidden from the screen — left the board."],[578.2255625,"honest is hidden from the screen — left the board."],[578.2255625,"readings2 is hidden from the screen — left the board."],[578.2255625,"row2 is hidden from the screen — left the board."]]},{"start":579.4255625000001,"say":"Written as a single chain, the whole proof is five lines. Zero does nothing, so the thing on the left is zero plus itself. That is rule two.","live":[],"does":[[579.4255625000001,"head_chain is shown on the screen, written out."],[580.5870625,"chain is shown on the screen, written out."]]},{"start":589.0350625,"say":"That zero is a plus the canceler of a: rule three. And now the only interesting line in the proof. Slide the bracket. Rule one.","live":["head_chain"],"does":[[590.3810625,"chain is shown on the screen, written out."],[596.4420625,"chain is shown on the screen, written out."],[598.4040625,"chain is indicated — a transient flash."]]},{"start":599.9295625,"say":"The middle and the right cancel, by rule three again, leaving a plus zero. And a plus zero is a, by rule two. Done.","live":null,"does":[[603.5230625,"chain is shown on the screen, written out."],[608.5730625,"chain is shown on the screen, written out."]]},{"start":610.0550625,"say":"Now notice what never happened in that chain. We never swapped two terms. The fourth rule, that a plus b equals b plus a, was never used.","live":null,"does":[[615.9070625,"p4 is shown on the screen, written out."],[619.1920625,"p4 is struck through — it is ruled out."],[619.7661318199213,"p4_note is shown on the screen, written out."]]},{"start":620.8605625,"say":"So this is not really a fact about numbers. It holds anywhere there is a zero, an undo for every move, and a bracket that slides. Rotating a cube. Shuffling a deck. The undo of the undo is the move you started with.","live":["p4","p4_note","head_chain"],"does":[[628.2090625,"chain is indicated — a transient flash."],[636.3245625,"chain is hidden from the screen — left the board."],[636.3245625,"head_chain is hidden from the screen — left the board."],[636.3245625,"p4 is hidden from the screen — left the board."],[636.3245625,"p4_note is hidden from the screen — left the board."]]},{"start":637.5245625,"say":"So back to the argument. Who was right? Both of them, describing one theorem from opposite ends.","live":[],"does":[[637.5245625,"head_who is shown on the screen, written out."],[637.5245625,"label_claim is shown on the screen, written out."],[641.4490625,"recipe is shown on the screen, written out."]]},{"start":645.4395625,"say":"The person who asked had the shape of it exactly right. a cancels minus a. Cancelers are unique. Therefore a is the canceler of minus a, and minus minus a is only its name.","live":["label_claim","recipe","head_who"],"does":[[649.5960625,"recipe (the \"Free: that is P3\" part) is emphasized."],[652.2540625,"recipe (the \"Cancelers are unique\" part) is emphasized."],[652.2540625,"recipe (the \"Free: that is P3\" part) is no longer emphasized."],[658.2330625,"recipe (the \"Cancelers are unique\" part) is no longer emphasized."],[658.2330625,"recipe (the \"its name\" part) is emphasized."]]},{"start":659.5995625,"say":"The objection was right too: you cannot get that middle line by notation. Which is exactly why the small proof earns its keep. It proves the middle line on the spot, with one slide of the bracket. The objector saw that, and said so. That is a good thread.","live":null,"does":[[659.5995625,"label_thread is shown on the screen, written out."],[663.1530625,"recipe (the \"Cancelers are unique\" part) is emphasized."],[663.1530625,"recipe (the \"its name\" part) is no longer emphasized."],[663.8490625,"voice_obj is shown on the screen, written out."],[674.3210624999999,"voice_end is shown on the screen, written out."],[675.9240625,"recipe (the \"Cancelers are unique\" part) is no longer emphasized."],[676.8476041666667,"head_who is hidden from the screen — left the board."],[676.8476041666667,"label_claim is hidden from the screen — left the board."],[676.8476041666667,"label_thread is hidden from the screen — left the board."],[676.8476041666667,"recipe is hidden from the screen — left the board."],[676.8476041666667,"voice_end is hidden from the screen — left the board."],[676.8476041666667,"voice_obj is hidden from the screen — left the board."]]}]},{"title":"What It Does Not Prove","start":677.8892708333333,"end":807.1205208333333,"objects":{"ask2":"a Panel that says \"Does this prove that a negative times a negative is a positive?\"","chips":"a Block [text] that says \"Free, by P3: $a$ cancels $-a$. Unique, by P1: nobody else does.\"","final_claim":"a Math [text] that says \"$-(-a) = a$\"","first":"a Polygon [yellow] drawn in floors (vertices=((0.6, 2.1), (5.4, 2.1), (5.4, 3.6), (0.6, 3.6)), fill_opacity=0.2)","first_tag":"a Point [yellow] labelled \"upright(\"multiplication\")\" drawn in floors (location=(3.0, 2.5), show_marker=False)","floors":"a Figure (x_range=(0.0, 6.0), y_range=(0.0, 6.0), aspect=(4.0, 4.2))","ground":"a Polygon [blue] drawn in floors (vertices=((0.6, 0.5), (5.4, 0.5), (5.4, 2.0), (0.6, 2.0)), fill_opacity=0.25)","ground_tag":"a Point [blue] labelled \"upright(\"addition\")\" drawn in floors (location=(3.0, 0.9), show_marker=False)","head_close":"a Heading that says \"One Partner Each\"","mtag":"a Tex [text] that says \"So $(-1) dot.op a$ cancels $a$. By uniqueness, $(-1) dot.op a = -a$.\"","mtag2":"a Math [text] that says \"$(-a)(-b) = a b$\"","mwork":"a Derivation [text] that says \"$a + (-1) dot.op a &= 1 dot.op a + (-1) dot.op a \\ &= (1 + (-1)) dot.op a = 0 dot.op a = 0$\"","p_a":"a Point [blue] labelled \"a\" drawn in pairing (location=(1.1, 3.0))","p_b":"a Point [blue] labelled \"b\" drawn in pairing (location=(4.3, 3.0))","p_edge_a":"a Line [green] drawn in pairing (start=(1.1, 3.0), end=(2.7, 3.0))","p_edge_b":"a Line [green] drawn in pairing (start=(4.3, 3.0), end=(5.9, 3.0))","p_loop":"a Circle [green] drawn in pairing (center=(3.5, 1.2), radius=0.35)","p_na":"a Point [yellow] labelled \"-a\" drawn in pairing (location=(2.7, 3.0))","p_nb":"a Point [yellow] labelled \"-b\" drawn in pairing (location=(5.9, 3.0))","p_zero":"a Point [text] labelled \"0\" drawn in pairing (location=(3.5, 1.2))","pairing":"a Figure (x_range=(0.0, 7.0), y_range=(0.0, 4.0), aspect=(5.0, 3.0))","second":"a Polygon [gray] drawn in floors (vertices=((0.6, 3.7), (5.4, 3.7), (5.4, 5.2), (0.6, 5.2)), fill_opacity=0.2)","second_tag":"a Point [gray] labelled \"upright(\"order\")\" drawn in floors (location=(3.0, 4.1), show_marker=False)"},"beats":[{"start":677.8892708333333,"say":"One more question from the thread, and it is a good one. Does everything we have just done prove that a negative times a negative is a positive?","live":[],"does":[[677.8892708333333,"ask2 is shown on the screen, written out."]]},{"start":686.3492708333333,"say":"No. And seeing why is worth a minute. Think of the rules as a building. Everything today happened on the ground floor: addition, zero, cancelers, and the bracket that slides.","live":["ask2"],"does":[[690.8542708333333,"floors is shown on the screen, written out."],[693.4662708333333,"ground is shown on the screen, written out."],[694.8482708333332,"ground_tag is shown on the screen, written out."]]},{"start":700.4522708333333,"say":"On this floor there is no multiplication at all, and no such thing as a positive or a negative number. Minus a is only the canceler of a, and it might be five, or minus five, or zero.","live":["floors","ask2","ground","ground_tag"],"does":[[701.1952708333333,"ground is indicated — a transient flash."]]},{"start":714.1827708333333,"say":"To even say negative times negative you need the next floor up: multiplication, and the rule that ties it to addition. Watch what happens there. Add a to minus one times a, and rewrite the first a as one times a.","live":null,"does":[[717.4222708333333,"first is shown on the screen, written out."],[718.6292708333333,"first_tag is shown on the screen, written out."],[727.1162708333333,"floors moves to a new place on the board."],[727.1162708333333,"mwork is shown on the screen, written out."]]},{"start":730.9382708333333,"say":"Pull the a out. One plus minus one is zero, and zero times anything is zero. So minus one times a cancels a. And because cancelers are unique, minus one times a is minus a. That is our engine, borrowed.","live":["floors","ask2","ground","ground_tag","first","first_tag"],"does":[[731.2572708333332,"mwork is shown on the screen, written out."],[741.3812708333332,"mtag is shown on the screen, written out."],[745.7002708333333,"mtag (the \"uniqueness\" part) is indicated — a transient flash."]]},{"start":747.6697708333332,"say":"From there, minus a times minus b works out to a times b, and that step uses today's theorem. But you still have not said the word positive. For that you need a third floor, the rules for which numbers count as positive, and only there does the sentence mean anything.","live":["mtag","floors","ask2","ground","ground_tag","first","first_tag"],"does":[[750.2932708333333,"mtag2 is shown on the screen, written out."],[757.8172708333333,"second is shown on the screen, written out."],[760.1732708333333,"second_tag is shown on the screen, written out."]]},{"start":765.4642708333333,"say":"Today's theorem is one brick in that wall. A load bearing one. But it is not the wall.","live":["mtag","mtag2","floors","ask2","ground","ground_tag","first","first_tag","second","second_tag"],"does":[[767.1362708333334,"ground is indicated — a transient flash."],[771.7567708333333,"ask2 is hidden from the screen — left the board."],[771.7567708333333,"floors is hidden from the screen — left the board."],[771.7567708333333,"ground is hidden from the screen — floors left the board."],[771.7567708333333,"ground_tag is hidden from the screen — floors left the board."],[771.7567708333333,"first is hidden from the screen — floors left the board."],[771.7567708333333,"first_tag is hidden from the screen — floors left the board."],[771.7567708333333,"second is hidden from the screen — floors left the board."],[771.7567708333333,"second_tag is hidden from the screen — floors left the board."],[771.7567708333333,"mtag is hidden from the screen — left the board."],[771.7567708333333,"mtag2 is hidden from the screen — left the board."],[771.7567708333333,"mwork is hidden from the screen — left the board."]]},{"start":772.9567708333333,"say":"So here is the picture to keep. Draw a line between two numbers whenever they add to zero. Rule three says every number gets at least one such line. Rule one says no number gets two. So the lines make a perfect pairing, and zero is its own partner.","live":[],"does":[[772.9567708333333,"head_close is shown on the screen, written out."],[775.6732708333333,"pairing is shown on the screen, written out."],[776.3002708333333,"p_a is shown on the screen, written out."],[776.580163517274,"p_na is shown on the screen, written out."],[776.7910562012147,"p_edge_a is shown on the screen, written out."],[781.2812708333333,"p_b is shown on the screen, written out."],[781.5417599562376,"p_nb is shown on the screen, written out."],[781.8928321468985,"p_edge_b is shown on the screen, written out."],[789.3272708333333,"p_zero is shown on the screen, written out."],[789.9865236642642,"p_loop is shown on the screen, written out."]]},{"start":791.0702708333333,"say":"Minus minus a equals a says exactly this: go to your partner, then go to your partner's partner, and you are home.","live":["pairing","head_close","p_a","p_na","p_edge_a","p_b","p_nb","p_edge_b","p_zero","p_loop"],"does":[[794.8262708333333,"pairing moves to a new place on the board."],[794.8262708333333,"final_claim is shown on the screen, written out."],[797.1362708333334,"p_a is indicated — a transient flash."],[798.4482708333333,"p_na is indicated — a transient flash."]]},{"start":801.0337708333333,"say":"Turn around twice: true. Why it is true: one slide of a bracket.","live":["final_claim","pairing","head_close","p_a","p_na","p_edge_a","p_b","p_nb","p_edge_b","p_zero","p_loop"],"does":[[801.7652708333333,"chips is shown on the screen, written out."],[803.4132708333333,"A box is drawn around final_claim."],[804.7372708333334,"chips (the \"Unique, by P1\" part) is emphasized."],[805.8288541666666,"chips (the \"Unique, by P1\" part) is no longer emphasized."],[806.0788541666666,"chips is hidden from the screen — left the board."],[806.0788541666666,"final_claim is hidden from the screen — left the board."],[806.0788541666666,"head_close is hidden from the screen — left the board."],[806.0788541666666,"pairing is hidden from the screen — left the board."],[806.0788541666666,"p_a is hidden from the screen — pairing left the board."],[806.0788541666666,"p_na is hidden from the screen — pairing left the board."],[806.0788541666666,"p_edge_a is hidden from the screen — pairing left the board."],[806.0788541666666,"p_b is hidden from the screen — pairing left the board."],[806.0788541666666,"p_nb is hidden from the screen — pairing left the board."],[806.0788541666666,"p_edge_b is hidden from the screen — pairing left the board."],[806.0788541666666,"p_zero is hidden from the screen — pairing left the board."],[806.0788541666666,"p_loop is hidden from the screen — pairing left the board."]]}]}]},"durationSeconds":807,"chapters":[{"title":"Two Minuses, and an Argument","startSeconds":0,"narration":"Everybody knows this one. Turn around, then turn around again, and you are facing the way you started. Flip a coin twice and it shows the same face. Cancel somebody's debt, and you have handed them money. Two minus signs make a plus. So why did one short question about it turn into a twelve hour argument, on a forum full of people learning mathematics? An argument that ended only when one of the two people in it wrote, you are right, my apologies. The two sides went like this. One of them said the fact holds in any system where you can add and cancel, so there is nothing to prove. The other said that the proof on offer assumed the very thing it was trying to prove. You cannot get a theorem by writing a symbol down in a suggestive way. They were both pointing at something real. There is a small, genuine theorem hiding inside this obvious fact, and the argument was about where it lives. To find it we have to work only from the rules. The person who asked was reading the first chapter of Spivak's Calculus, which opens by writing down the rules that numbers obey. From here on we are allowed exactly what the rules say and nothing else. No obviously. No number line. Rule one: grouping does not matter. If you are adding three things, you may bracket the first pair or the last pair, and you get the same answer. Rule two: there is a number called zero, and adding it does nothing at all. And rule three, the one the whole argument is about. For every number a, there is a number, written minus a, that cancels it. a plus minus a is zero, and minus a plus a is zero as well. Now read that third rule again, slowly, in words. Four of those words are going to carry this entire video. There is a number. It says there is a number. It never says there is only one. And notice one last thing before we go on. The fourth rule on that list, that a plus b equals b plus a, we are never going to use."},{"title":"The Minus Sign Is a Job Title","startSeconds":130.90145833333332,"narration":"Three rules and a claim. Before we argue about the claim, we had better be sure what its symbols mean, and the minus sign is where people slip. In this rulebook the minus sign is not a property of a number. It is not negativeness. It is a job title. Minus a means whichever number holds the job of cancelling a. Rule three hands us that number and says what the job is: a plus minus a is zero. Let me draw it. Here is a, here is its canceler, and I will join two numbers with a line whenever they add to zero. If a is five, the canceler is minus five. If a is minus three, the canceler is three, so a symbol with a minus in front can perfectly well be a positive number. And zero cancels itself. Now read the thing we are trying to prove from the inside out. Minus a is the canceler of a. So minus minus a is the canceler of the canceler of a. I am keeping that one grey, with a question mark on it, because at this point in the story we do not know which number it is. Here is where the thread got tangled, and it is a good tangle. Does a cancel minus a? Yes, at once. Rule three says minus a plus a is zero, and that is exactly the job. So a does it, for free. So are we done? Not quite, and the difference is a single word. What we have shown is that a is a canceler of minus a. What the symbol names is the canceler of minus a. Look at rule three once more. It promised that a canceler exists. It never promised there is only one. If minus a had two different cancelers, the symbol would be ambiguous. It might be a. It might be the grey one. Two minuses making a plus would be a coin toss. So here is the theorem underneath the claim. Every number has exactly one canceler. Once you have that, minus minus a equals a is only that theorem, read out at the number minus a. And you might think exactly one canceler is obviously true. It is not."},{"title":"A World Where It Fails","startSeconds":273.97304166666663,"narration":"To see that exactly one canceler is not obvious, let me build a number system where it is false. Four numbers: zero, a, b and c. Its addition is given by a table, the same kind you learned your sums from, only smaller. Check rule two. The zero row and the zero column hand everything straight back, unchanged. Zero does nothing here. Check rule three. Every number here has a canceler. a and b add to zero. c and b add to zero. And zero cancels itself. The table is even symmetric, so this addition commutes too. Now look at the row for b. It has two zeros in it. b is cancelled by a, and b is also cancelled by c. In this world, b has two cancelers. So compute the canceler of the canceler of a. The row for a has exactly one zero, in the b column, so minus a is b. Now the row for b has two zeros, so the canceler of b is a, or it is c. Take c. In this little world, minus minus a is c, and c is not a. Which rule did we break? Not rule two. Not rule three. And commutativity is fine as well. Look at rule one. Add a and b first, then add c: a plus b is zero, and zero plus c is c. Now bracket the other pair: b plus c is zero, and a plus zero is a. And look at the two values we just got. The first was c. The second was a. Those are exactly the two numbers that were fighting over the title canceler of b. That is not a coincidence. It is our theorem, seen from the wrong side."},{"title":"The Bracket Slides","startSeconds":382.5438958333333,"narration":"So what does our number system have that the toy one does not? One thing. The bracket can slide. Here is what that buys us. Suppose two numbers, a and c, both cancel the same number b. So a plus b is zero, and b plus c is zero. I am assuming nothing else about them. Must a and c be the same number? Write all three of them in a row: a plus b plus c. Rule one says this row has one value, no matter how I bracket it. So let us bracket it two different ways and compare. Bracket the left pair first. a and b cancel, so that pair is zero. Zero plus c is c. The whole row is worth c. Now the one motion to remember from this whole video. Watch the bracket leave the left pair, and land on the right pair instead. Now the same thing happens at the other end. b and c cancel, so that pair is zero. a plus zero is a. The row is worth a. Same row. One value. So a equals c, and that is the proof. Any two cancelers of the same number are equal. Cancelers are unique. There is nothing clever in that argument. a and c only had to be in the same room as b, and rule one is what puts them there. In the toy world the bracket cannot slide, so a and c never have to meet, and nothing makes them equal. That is the only difference between the two worlds. Let us write that down as a theorem, because we are about to use it twice. If a cancels b, and c also cancels b, then a and c are the same number. Rule three says a canceler exists. Rule one says there is only ever one."},{"title":"Cashing It In","startSeconds":502.8800625,"narration":"Before we cash it in, let us fix what the symbol is allowed to mean. Minus minus a is a canceler of minus a, any one of them. We are not assuming there is only one of them, because that is the thing we are proving. Now rule three gives us two promises. First: a has a canceler, and the two of them add to zero. Second: minus a has a canceler too, and minus minus a is the name of one. Write all three in a row, exactly as before: a, then the canceler of a, then the canceler of that. Rule one says this row has one value however I bracket it. Bracket the left pair. a and its canceler make zero, and zero plus the last slot leaves the last slot standing. So the row is worth minus minus a. Now slide it. The bracket leaves the left pair and lands on the right pair instead. And now the other two cancel, by the promise we just wrote down. a plus zero is a. So the row is worth a. Same row, one value. So minus minus a equals a. That is the theorem we came for, and it is now proved. Written as a single chain, the whole proof is five lines. Zero does nothing, so the thing on the left is zero plus itself. That is rule two. That zero is a plus the canceler of a: rule three. And now the only interesting line in the proof. Slide the bracket. Rule one. The middle and the right cancel, by rule three again, leaving a plus zero. And a plus zero is a, by rule two. Done. Now notice what never happened in that chain. We never swapped two terms. The fourth rule, that a plus b equals b plus a, was never used. So this is not really a fact about numbers. It holds anywhere there is a zero, an undo for every move, and a bracket that slides. Rotating a cube. Shuffling a deck. The undo of the undo is the move you started with. So back to the argument. Who was right? Both of them, describing one theorem from opposite ends. The person who asked had the shape of it exactly right. a cancels minus a. Cancelers are unique. Therefore a is the canceler of minus a, and minus minus a is only its name. The objection was right too: you cannot get that middle line by notation. Which is exactly why the small proof earns its keep. It proves the middle line on the spot, with one slide of the bracket. The objector saw that, and said so. That is a good thread."},{"title":"What It Does Not Prove","startSeconds":677.8892708333333,"narration":"One more question from the thread, and it is a good one. Does everything we have just done prove that a negative times a negative is a positive? No. And seeing why is worth a minute. Think of the rules as a building. Everything today happened on the ground floor: addition, zero, cancelers, and the bracket that slides. On this floor there is no multiplication at all, and no such thing as a positive or a negative number. Minus a is only the canceler of a, and it might be five, or minus five, or zero. To even say negative times negative you need the next floor up: multiplication, and the rule that ties it to addition. Watch what happens there. Add a to minus one times a, and rewrite the first a as one times a. Pull the a out. One plus minus one is zero, and zero times anything is zero. So minus one times a cancels a. And because cancelers are unique, minus one times a is minus a. That is our engine, borrowed. From there, minus a times minus b works out to a times b, and that step uses today's theorem. But you still have not said the word positive. For that you need a third floor, the rules for which numbers count as positive, and only there does the sentence mean anything. Today's theorem is one brick in that wall. A load bearing one. But it is not the wall. So here is the picture to keep. Draw a line between two numbers whenever they add to zero. Rule three says every number gets at least one such line. Rule one says no number gets two. So the lines make a perfect pairing, and zero is its own partner. Minus minus a equals a says exactly this: go to your partner, then go to your partner's partner, and you are home. Turn around twice: true. Why it is true: one slide of a bracket."}]}}