# I Can you use proof by contradiction in the midst of induction

1. Mar 21, 2016

### Battlemage!

In the process of doing a proof by induction, can you use a contradiction to show that if P(k) holds then P(k+1) must hold? What I mean is, after establishing that P(0) holds, can I assume that P(k) holds and that P(k+1) does not, and show that a contradiction arises, and thus conclude that if P(k) holds then P(k+1) must also hold? If I can, then in combination with showing that P(0) holds, then the proof for induction would be complete if I understand the process.

Let me give an example.

Say I wanted to prove that if (u1)2, (u2)2, ..., (un)2 are all each ≥ 0 for u1, u2, ... un in R,
then the sum (u1)2+ (u2)2+ ...+ (un)2 ≥ 0.

Well, clearly (u1)2 and (u2)2 are each ≥ 0 by the rules of squared numbers in R, so it's clear that (u1)2+ (u2)2 ≥ 0. So let's say I have shown that (I'm sure to actually prove it I'd have to get into some basic axioms, but let's just pretend that has been done).

Usually in an inductive proof, if I understand it correctly, the next step would be to assume that
(u1)2+ (u2)2+ ...+ (uk)2 ≥ 0 and use that to show that it therefore follows that (u1)2+ (u2)2+ ...+ (uk)2 + (uk+1)2 ≥ 0.
Can I instead assume that (u1)2+ (u2)2+ ...+ (uk)2 ≥ 0 AND assume that (u1)2+ (u2)2+ ...+ (uk)2 + (uk+1)2 < 0,
and show that I arrive at a contradiction, thereby proving that P(k) => P(k+1)? (Essentially what I'd be doing is instead of directly showing that if P(k) holds then P(k+1) holds, I'd be showing that if P(k) holds and P(k+1) does not hold that a contradiction is reached, and therefore if P(k) holds then P(k+1) holds).

So it would look like this:
Assume that (u1)2+ (u2)2+ ...+ (uk)2 ≥ 0 holds and that (u1)2+ (u2)2+ ...+ (uk)2 + (uk+1)2 < 0.

Then that would mean that
(u1)2+ (u2)2+ ...+ (uk)2 < -(uk+1)2.
By the same reasoning I would have used to show that (u1)2 and (u2)2 ≥ 0, I would show that (uk+1)2 ≥ 0, which would make - (uk+1)2 < 0, but that would mean that (u1)2+ (u2)2+ ...+ (uk)2 < 0, which would contradiction my assumption that (u1)2+ (u2)2+ ...+ (uk)2 ≥ 0. By contradiction, it would have to mean that the case for P(k+1) holds.

Granted, this proof outline itself is garbage and more or less ignores that a number times itself is positive in R, but I'm more worried about the principle itself. After showing P(0) is true, can we show that if P(k) holds then P(k+1) holds by assuming it does NOT and then arriving at a contradiction?

Also, if you want to critique/correct my "fake" proof please do. Thanks and sorry for this weird and all over the place question.

2. Mar 21, 2016

### andrewkirk

Yes you can do that.
$P(k)\wedge \neg P(k+1)\Rightarrow \bot$ is logically equivalent to $P(k)\Rightarrow P(k+1)$, by the use of the inference rules
- Reductio Ad Absurdam (RAA); and
- Conditional Proof (CP)

which you can see in this list of inference rules.

3. Mar 21, 2016

### Red Goose

Sure. The law of contradiction is fundamental. So long as you are using it properly (for instance, you must actually prove the contrary to be impossible, not disprove one counter-example) then you're good to go.

You don't need the rules of squared numbers in R, we get that fact simply from your assumption. You don't need to prove your antecedent true.

This all looks great. Well done. Only thing I would suggest you change (and its a minor detail) is to specify that you are doing a proof by contradiction in your declarative statement. Instead of saying "Assume that (u1)2+ (u2)2+ ...+ (uk)2 ≥ 0 holds and that (u1)2+ (u2)2+ ...+ (uk)2 + (uk+1)2 < 0." say "Assume by way of contradiction that (u1)2+ (u2)2+ ...+ (uk)2 ≥ 0 holds and that (u1)2+ (u2)2+ ...+ (uk)2 + (uk+1)2 < 0."

4. Mar 22, 2016

### Battlemage!

Thanks for the responses. I was a little iffy on that probably because induction isn't my strong suit. Also I am glad to know I can state exactly what I'm doing without losing the proper language of a proof (something I am definitely still learning).

Incidentally I just realized my "test proof" is a proof that the dot product of a vector u and itself is always greater than or equal to zero. Because obviously each component times itself and then summed up over all all the indices would look exactly like the above.