# Direct proof confusion?

1. Sep 6, 2012

### halo31

1. The problem statement, all variables and given/known data
im supposed to use a direct proof to prove that if 1-n^2>0 then 3n-2 is odd for all n∈Z

2. Relevant equations

3. The attempt at a solution
if you let n∈z then suppose that 1-n^2>0 then 1>n^2 but the only inter n such that 1>n^2 is 0. 3x0-2=-2 as -2=2(-1), -2 is even . Hence if 1-n^2>0 then 3n-2 is even when n∈z. The part that im confused on is why in the book does it say for directs proofs to choose an arbitrary value for n and prove that Q(X) is true? In this proof I had to prove it the same way but there was no way I could let N represent an arbitrary value. The only value that worked was when n=0.

2. Sep 6, 2012

### christoff

The statement you're trying to prove does not make sense. You're saying you have the prove this:
$1-n^2>0 \Rightarrow 3n-2$ is odd, for ALL $n\in\mathbb{Z}$.

What about n=0? Then 1-n^2=1>0, so the left side is satisfied. If this were true, then you would have that 3n-2=-2 is odd. But it isn't.

Moreover, if you're talking about integers, then the only integer which satisfies your left side inequality is n=0, which we've just shown doesn't even work.

In conclusion... this statement is just wrong. Wrong, wrong, wrong.

3. Sep 6, 2012

### halo31

my bad not for all nεZ just that n is an element of Z

4. Sep 6, 2012

### christoff

Please copy out the question directly from your textbook, with all the punctuation. I still think you've copied down the question wrong, because n=0 is the only integer for which this holds.

5. Sep 6, 2012

### halo31

Its "Let nεZ, Prove that if 1-n^2>0, then 3n-2 is an even integer" That's all it give me and yes 0 is the only integer that satifies 1-n^2>0. But my question is that how can this be considered a direct proof when you have to assume that n represents some arbitrary value in the domain or is this not always the case?

6. Sep 6, 2012

### christoff

It's a bit awkward, but you can prove it directly...

Direct proof:

Claim 1: n=0 satisfies the inequality.
Proof: this is trivial.
Claim 2: n=0 is the only integer which satisfies that inequality.
Proof: Since $n^2>0$ for all $n\neq 0$, we need only consider positive integers. Then for any n>0, we have $n^2\geq 1$, so $1-n^2\leq 0$. Hence, n=0 is the only integer which satisfies the inequality.
Claim 3: n=0 satisfies "3n-2 is even"
Proof: Trivial.

This is a direct proof because at no point along the way did I assume a "false" hypothesis in order to come to a contradiction. It may seem as though Claim 2 was done by contradiction, but it wasn't. I am claiming for all $n\neq 0$, we do NOT have $1-n^2>0$, equivalently, we DO have $1-n^2\geq 0$.

Last edited: Sep 6, 2012
7. Sep 6, 2012

### halo31

dont you mean for the last part of claim 2 to be 1-n^2≥0?

8. Sep 7, 2012

### christoff

Do you mean the claim, or the proof of the claim? In either case, no, I do not. I'm attempting to prove that for all nonzero integers, the inequality 1-n^2>0 does NOT hold. What does it mean for an inequality to NOT hold? It means that the opposite inequality DOES hold. That is, 1-n^2≤0 holds.

9. Sep 7, 2012

### HallsofIvy

Staff Emeritus
The crucial point is that the only integer for which "1- n^2> 0" is n= 0.

If n= 1 or -1, what is 1- n^2? If n= 2 or -2, what is 1- n^2? If n= 3, or -3, ...

Personally, I would have started this proof with "If n is an integer such that 1- n^2> 0, then n= 0. Therefore, 3n-2= ....

10. Sep 7, 2012

### halo31

Well I reread the section in the book on direct proofs and I happen to gloss otver the part when it said
'''in order to prove for all x in the domain of S'', and I assumed you needed to let any variable such as x,n etc to represent an arbritrary value in the given domain. Now I know that I need to pay attn. to what values in the domain satisfy the hypothesis.