Which school algebra identities survive the enlargement, and why

0:00An identity is not a sum. It is a promise about every sum of a certain shape. The square of a sum equals the sum of the squares plus twice the product. That says nothing about any particular pair. It says: whichever two you pick, this holds. Which raises a question everyone skips. Pick them from WHERE? The system just got bigger. There are numbers in it that were not there when you learned that identity.

0:30And your reason for believing it was that it is true for ordinary numbers. That reason no longer covers the cases you care about. So: prove every identity again for the new numbers. Or assume they still work and hope. Or look at what one proof actually uses, and see how far that reaches. Only the third scales, and it is the one this video does. Start with the identity for the square of a sum, both sides written out.

1:03On the left, a bracket multiplied by itself. On the right, three terms: a square, twice a product, another square. The claim is that these two are the same number for every pair in the enlarged system. Not for a pair you tried. For every pair. And a claim about every pair cannot be settled by trying pairs. It has to be derived. Derived from what? That turns out to be the whole answer.

1:34The enlarged system was granted five laws. The order of a sum. The grouping of a sum. The order of a product. The grouping of a product. And spreading a product over a sum. Nothing else was granted. So nothing else may be used. Here is the derivation, as a ladder. At the top, the square, written honestly as the bracket times itself. First rung: spread the whole bracket over the two terms inside the second one. That gives a sum of two products.

2:10Second rung: spread the left of those two. Now you have the first term's square, and a cross term. Third rung: spread the right one as well. Four terms. Four, not two - which is already the answer to the commonest wrong expansion. Fourth rung, and this is the one to watch. The two middle terms are written in opposite orders. Swap one, and they become the same thing. Fifth and sixth rungs: regroup so the two middles sit together, and there is your twice-the-product.

2:44Six rungs. Every one of them small enough that a single law permits it. Now take each rung and ask, of all five granted laws, which ones permit it. Not which one was written beside it. Which ones actually work. Rungs one, two and three: spreading, and only spreading. Rung four: the order of a product, and only that. Rungs five and six: the grouping of a sum. No rung is permitted by two different laws. Every step in this proof has exactly one licence.

3:21Count them up and the proof spends three rungs on spreading, two on the grouping of a sum, and one on the order of a product. Which leaves two of the five granted laws carrying nothing at all. This proof never once needs the order of a sum, and never once needs the grouping of a product. They are not idle laws. They are simply not what this argument is made of, and knowing that will matter shortly.

3:51That table was found by trying. Check it by withdrawing a law and running the same ladder again. Withdraw spreading, and the first three rungs have no licence left at all. The last three are untouched. Withdraw the order of a product, and rung four dies. Only rung four. Withdraw the order of a sum, and nothing happens. Every rung still stands. That is what it means to say a line uses a law: take the law away and watch which line falls.

4:25And notice what it does not mean. Rung four losing its licence does not prove there is no other route to the same place. A search that comes back empty means not on this list. It never means impossible. So if the order of a product is really doing work here, something stronger is owed. Here is the stronger thing. Build a different system. Two slots again, added the same way, but multiplied by a rule chosen to obey every granted law except the order of a product.

5:03In that system, is the identity still true? First, check the expansion itself survives. Spread the bracket twice and you still get four terms, at every one of the two thousand four hundred and one pairs swept. But the two middle terms are the same thing at only two hundred and seventy-one of them. At the other two thousand one hundred and thirty they are different numbers - and the identity is false at exactly those. The pairs where it fails are the pairs where the middles refuse to combine.

5:39Take one and the symbol. Their sum squared is one plus the symbol. The identity predicts one plus twice the symbol. Note what did not happen. The middle terms did not cancel, and the symbol did not make them vanish. They stayed. They simply could not be added together. Four more identities usually follow with no proof at all. The square of a difference. The cube of a sum. The cube of a difference. And a difference of two squares.

6:14That is not laziness. It is the point. Look again at the ladder. Nothing in it depends on the two letters being letters. Put a product in place of the second and run the same six rungs. Every rung is permitted by the same law as before. Put a whole sum in place of the first as well. The expressions grow from seven parts to eleven to fifteen, and the licences do not change by a single entry.

6:44So the ladder was never about x and y. It was about the SHAPE. Which is why the square of a difference needs no new argument: a difference is a sum, with the negative of the second put in. The proof you have read covers it, and the cubes, and the difference of two squares. One derivation, five identities. That is what a proof buys and a hundred checked examples cannot.

7:14To see how tightly the laws control the identities, put six multiplication rules side by side. The one that is used. Slot by slot. The minus forgotten. One that ignores the right-hand second slot. One with its answers swapped. One that throws its last term away. For each, two questions. Which granted laws does it break, and which of the five identities does it fail? Three of the six break no law. Those three fail no identity - not one, across all five.

7:51The rule that loses only the order of a product fails all five. So does the one that loses that and the grouping too. And then the interesting one. The rule with its answers swapped keeps the order of a product but loses the grouping. It fails the two cubes and nothing else. Both squares survive there, and so does the difference of two squares. Which is what the tally predicted. A cube must be read from one end or the other, and that reading is the grouping of a product - a law the square's ladder never touches.

8:29Break a law and you break precisely the identities whose proofs use it. Not more, not fewer. Now the stress test. Go back through the six rungs and look for anything only ordinary numbers can do. There is no division anywhere in it. No square roots. Nothing is ever compared as larger or smaller than anything else. And at no point does the argument lean on a square being positive.

9:01It is rearrangement, from top to bottom. Nothing else. One more way to see it. Leave the rule alone and change what the symbol squares to - all nine whole numbers from minus four to four. Across all nine, the five identities fail zero times. Not once. And these really are nine different systems - two of them already disagree on one hundred and forty-four of two hundred and fifty-six products.

9:32The identities cannot see the difference, because the proof never asks what the symbol squares to. So does everything true of ordinary numbers come across? No. And the reason is the same reason the identities do. The transfer reaches exactly as far as the granted laws, and one very ordinary thing was never granted: an order. A way of saying one number is bigger than another. You might think you can invent one. You can - that is the problem.

10:06Sort by the first slot, breaking ties with the second. A perfectly good total order. Or sort by the second slot first. Also perfectly good. Where the second slot is empty the two agree completely, so both look like the order you already know. Across the whole pool they disagree eight hundred and eighty-two times. Under either of them, some square is below nothing - twenty-three of them for one reading, twenty-four for the other. They do not even agree on how many.

10:41A question two honest readings answer differently is a question the laws do not settle. No argument built from those laws can reach it. The identities are not decoration. They are how you work. Take the cube of five minus three times the symbol. Multiplying the bracket out three times is possible and horrible. The cube identity is four terms. First cubed is one hundred and twenty-five. Three times the first squared times the second is two hundred and twenty-five of the symbol, entering with a minus.

11:23Three times the first times the second squared is fifteen times nine times the symbol squared - which is minus one hundred and thirty-five. And the second cubed brings back twenty-seven of the symbol. Collect: minus ten in the first slot, minus one hundred and ninety-eight in the second. Multiplying out three times gives the same, as it must. For a fourth power, do not look for a fourth-power identity. Square, then square again. One minus the symbol squares to minus twice the symbol, and that squares to minus four.

12:03A third plus three times the symbol cubes to minus two hundred and forty-two over twenty-seven, minus twenty-six of the symbol. One last use - the one you will reach for most. A difference of two squares says a sum times a difference is one square minus the other. Put the symbol into the second term. Two plus three of the symbol, times two minus three of the symbol. That is two squared minus three-of-the-symbol squared. Four minus nine times the symbol squared. Four plus nine.

12:42Thirteen. An ordinary number, with nothing in the second slot. An identity learned for ordinary numbers has turned a two-slot denominator into a one-slot one - because its proof only ever rearranged things. So the identities did not survive by luck, and they are not new ones that happen to look familiar. They are the same identities on a different foundation: no longer true because they hold for ordinary numbers, but derivable from laws this system was handed.

13:18Which is why you can tell in advance what will not come across. Read the proof. Whatever it uses, you need. Whatever it never touches cannot stop it.