Proving Evenness: 3a+1 is Even if and only if (a+1)/2 is an Integer

[hammonjj]

Homework Statement

Let a$\in$Z. Prove that 3a+1 is even if and only if (a+1)/2 $\in$Z

Homework Equations

We know that C is an odd number if there exists:

C=2k+1

Even:

C=2k

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...

 

[SammyS]

Homework Statement

Let a$\in$Z. Prove that 3a+1 is even if and only if (a+1)/2 $\in$Z

Homework Equations

We know that C is an odd number if there exists:

C=2k+1

Even:

C=2k

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

Therefore
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...

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.​

​

 

[hammonjj]

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.

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.

 

[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").

 

[hammonjj]

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").

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

 

[SammyS]

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

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?