# Prove a Number is Even

1. Feb 15, 2012

### hammonjj

1. The problem statement, all variables and given/known data
Let a$\in$Z. Prove that 3a+1 is even if and only if (a+1)/2 $\in$Z

2. Relevant equations
We know that C is an odd number if there exists:

C=2k+1

Even:

C=2k

3. The attempt at a solution
I think I figured it out, but I'm terrible at Discrete Math, so I was hoping for some input. We know that:

3a+1=2k

If we want 3a+1 to be even. Given the extra constraint we have, I said:

3[(a+1)/2]+1=2k

After some simplification:

(3a+5)/2

I then factored out a positive 2 and:

2[(3a+5)/4]

From here:
(3a+5)/4=k

Therefor
2k=2k

Am I way off base here or is this actually correct?

Thanks! I have a bunch of homework problems due at the end of the week, so please be patient with me! This class is ruining my life...

2. Feb 15, 2012

### SammyS

Staff Emeritus
Yes, you assumed what you were to prove.

This is an if and only if statement. You have to do both of the following:
1. Assuming that 3a+1 is even, show that (a+1)/2 is an integer.

From what you have above, you know that means:
Assume 3a+1 = 2k, where a and k are integers.

Show that (a+1)/2 = m for some integer m. (Is this saying that a+1 is even?) I underlined the word "show" because this should be a result. Don't assume this -- at least, not for this part.​
2. Assuming that (a+1)/2 is an integer, show 3a+1 is even.

From what you have above, you know that means:
Assume (a+1)/2 = n, where a and n are integers.

Show that 3a+1 = 2p, for some integer p.​

3. Feb 15, 2012

### hammonjj

How exactly do I show the (a+1)/2=m? I don't think I can assume that a+1 is even, since I have no basis for that conclusion, because, that would imply that a is odd. Or is that the point? Do I have to make a statement saying that the only way this is true is if a is odd? If so, how exactly do I write that?

My professor is a super harsh grader, so I want to make sure I do this 100% write. Most of the time, he doesn't even take points off for the proof itself, but the wording and formatting of the proof.

Last edited: Feb 15, 2012
4. Feb 15, 2012

### Deveno

we have two problems in one:

the first problem is, given that 3a+1 is even, we want to show that (a+1)/2 is an integer.

the second problem is, given that (a+1)/2 is an integer, we want to show that 3a+1 must be even.

one approach to problem "one":

if 3a+1 is even, 3a is odd.

if a were even, 3a would be even, so a is odd.

if a is odd, a+1 is even. therefore....

one approach to problem "two":

if (a+1)/2 is an integer, a+1 is divisible by 2.

thus a+1 = 2k, for some integer k (you could use any letter instead of k).

now write 3a+1 in terms of k:

3a + 1 = 3(2k - 1) + 1 = ...?

prove your result is an even integer (hint: "even" = "divisible by 2").

5. Feb 15, 2012

### hammonjj

I understand everything up until you rewrite 3a+1 in terms of of k. Why is there a 2k-1?

6. Feb 15, 2012

### SammyS

Staff Emeritus
I would say it a bit differently:

Once you have
a+1 = 2k ,​
I suggest that you add 2a to both sides of the equation.

Do you see how that will show that 3a+1 is even?