# Direct proofs help

1. Jun 12, 2012

### bonfire09

1. The problem statement, all variables and given/known data

Let nεZ,Prove that 1-n^2>0, then 3n-2 is an even integer.

2. Relevant equations

3. The attempt at a solution

I proved it like this. I think its right but im not able to word it correctly.

Since 1-n^2>0 therefore n=0. Then 3n-6=3(0)-6=-6. Since 0 is an integer, 3n-6 is even.

How can I learn to word this correctly because im having some trouble with it?

2. Jun 12, 2012

### Whovian

Try posting in the number theory forum, this isn't really calculus.

3. Jun 12, 2012

### bonfire09

this is intro to proofs actually. Im trying to self study.

4. Jun 12, 2012

### Whovian

5. Jun 12, 2012

### Reptillian

I wouldn't worry too much about proper wording as long as you get the concept. n has to equal zero and -6 is an even integer...sounds proved to me. :)

Although, you probably shouldn't take my advice. I'm shunned by many in academia due to my deep detestation of pretentiousness. ;)

6. Jun 12, 2012

### Staff: Mentor

The above is confusing. A better statement would be
Let n $\in$ Z. If 1 - n2 > 0, then show that 3n - 2 is an even integer.
Note that you have a typo in your work. You're supposed to prove that 3n - 2 is an even integer.

I would say it like this:
Since 1 - n2 > 0 and n $\in$ Z, then n = 0.
So 3n - 2 = - 2, which is an even integer.

Therefore, for any integer n, if 1 - n2 > 0, then 3n - 2 is an even integer.

It should NOT be posted in the number theory section. That section and the other sections under Mathematics are not for homework and homework-type problems

No, here is probably fine, although the Precalc Mathematics section would also be a good choice.