Proving It for Every Number

Last timeBreaking a Claim With One Case

Infinitely many cases proved by two finite checks: that the first case holds, and that each case forces the next. The second check is where nearly every broken argument of this kind breaks.

A claim about every whole number has infinitely many cases and cannot be

checked. The previous lesson said one bad case destroys such a claim; this one

is about establishing it, and it does so with two finite checks.

FIG 1What the two checks are

: the claim being made about the number n, such as the first n numbers adding to n(n+1)/2
: that claim with 1 put in, which is usually checked by direct arithmetic
: if-then, so the step is a conditional claim and not an assumption
: for every whole number n, from the first one onwards

The step is an if-then claim of exactly the kind from the third lesson, and establishing one never requires its condition to be true. That is what answers the objection everybody has on meeting this method, which is that assuming the claim for n looks like assuming what is to be proved. Nothing is assumed: what is shown is that the truth of one case would force the truth of the next, and that is a claim about the relationship between consecutive cases rather than about either one of them.

FIG 2The first n whole numbers add to n times n plus one, over two

step	stage	what is written	why it is allowed
1	the claim	1 + 2 + ... + n = n(n+1)/2	a claim with one case per whole number
2	the first case	left side is 1, right side is 1 times 2 over 2, which is 1	arithmetic, nothing more
3	set up the step	suppose the claim holds for some particular n	this is the condition of an if-then claim, not an assumption
4	the next case, left side	1 + 2 + ... + n + (n+1)	the sum one term longer
5	use the supposition	n(n+1)/2 + (n+1)	the first n terms replaced by what they equal
6	tidy	(n+1)(n+2)/2	factor out n plus one and add the halves
7	compare	that is the claim with n+1 in place of n	so the step holds, and with the first case the claim follows

7 steps

The fifth row is the only place the supposition is used, and in a well written proof of this kind there is exactly one such place, which is worth pointing at explicitly when writing one up. The sixth row is ordinary algebra with no logic in it. Note what the argument did not do: it never checked the claim for two, or ten, or a hundred. Those cases are covered because the step covers every n, and that is the whole economy of the method, two checks in place of infinitely many.

FIG 3Four arguments, scored on the two halves

	first case shown?	step sound?	claim actually true?
the sum formula above	1	1	1
every number equals the	0	1	0
every group of horses is	1	0	0
every number above 3 is	1	0	0
the two to the n growth	0	1	0

The second row has a perfectly sound step, since adding one to both sides of n equals n plus one gives the next case, and no first case, and the claim is false everywhere. That marked cell is why the first case is not a formality. The third row is the classic broken argument: the first case holds, a group of one horse being one colour, and the step fails exactly at the move from one to two, where it needs two overlapping groups and there is no overlap. The fourth row is a true-looking claim whose step nobody has ever produced; it is not known to be false and it is not proved, which is a different situation from the rows above it. The last row is a claim whose step only works once the number is large enough, so the chain never starts.

What goes wrong

The two defects to check for are the step that needs the number to be large

enough, and the step that uses more than the single case before it. The second

is repairable: there is a version of the method where the step may assume every

earlier case rather than only the previous one, and it is just as valid. But

the repair must be stated, because an argument claiming to use one case and

quietly using several is an argument nobody can check.

The lesson stops here

1 more paragraph to go

You have read the opening. The rest of the argument, the problems that check whether it landed, and the lines worth keeping at the end all come with a plan.

The first lesson of every course in the library reads the whole way through, free, so you can see exactly what the rest of them are.

See the plan The contents

This is the reading half

Starting the course gives you your own copy of it. Every idea on every page has problems standing under it, marked with a reason rather than a tick, and any sentence you do not believe can be opened and argued with. None of that can happen on a page nobody owns.

The contents

Proving It for Every Number

What goes wrong

1 more paragraph to go

This is the reading half

Read alongside