If n is an integer, and 3n+2 is even, prove that n is also even

[velouria131]

I am coming across hiccups in my proof process. I am given this problem - Prove: if n is an integer and 3n + 2 is even that n is also even. I have to apply a contrapositive proof to this problem. The form is then \negq therefore \negp .The problem becomes - if n is odd, prove that 3n+2 is even.

Work:

Prove - if n is odd, prove that 3n+2 is even.

step 1 - if n is odd, n = 2k+1 for some integer k

step 2 - 3n + 2 = 3(2k+1) + 2 = 6k + 5

step 3 - This is my issue. A contrapositive proof for this problem would give not 'p', or, that 3n+2 is odd when n is odd. Do I now have to show that 6k + 5 is an odd number for any positive integer k? Or, should I just prove that 3n + 2 is odd when n is odd? If I take this route, could I choose another proof method, essentially having a 'proof within a proof'?

 

[pwsnafu]

Prove: if n is an integer and 3n + 2 is even that n is also even. I have to apply a contrapositive proof to this problem. The form is then \negq therefore \negp .The problem becomes - if n is odd, prove that 3n+2 is even.

I'm assuming that's a typo.

Do I now have to show that 6k + 5 is an odd number for any positive integer k?

Probably not, but it's better to be safe. An odd number is an integer of form 2m+1. Find m and you are done.

 

[velouria131]

Probably not, but it's better to be safe. An odd number is an integer of form 2m+1. Find m and you are done.

I am still not sure where I would take this proof. I apologize in advance as this is maybe the third proof I have done, and lack serious intution. How would I go about asserting that 6k + 5 is odd for any integer k? Would I do this:

6k + 2 = 2m + 1

m = 3k + 2

...however, this feels like circular logic. Does this mean that 6k+2 takes the form of an odd integer, and is therefore odd?

 

[rollingstein]

What about this?

3n+2=2k where k is some integer

n=2(k-1)/3

If n is an integer (given) it has to be even with 2 as a factor.

 

[pwsnafu]

I am still not sure where I would take this proof. I apologize in advance as this is maybe the third proof I have done, and lack serious intution. How would I go about asserting that 6k + 5 is odd for any integer k? Would I do this:

6k + 2 = 2m + 1

m = 3k + 2

...however, this feels like circular logic. Does this mean that 6k+2 takes the form of an odd integer, and is therefore odd?

You need to stop being careless with your posting. It's "+5" not "+2".
And yes. We prove something is odd by either

1. Showing that it is equal to 2m+1 for some integer m, or
2. Show the number is congruent to 1 (mod 2).

And it's easy enough to show that those two statements amount to the same thing.

What about this?

3n+2=2k where k is some integer

n=2(k-1)/3

If n is an integer (given) it has to be even with 2 as a factor.

That's not a contrapositive proof.

 

[rollingstein]

That's not a contrapositive proof.

Sorry. Didn't read that requirement. My bad.

 

[rollingstein]

P1:If 3n+2 is even then n is also even

P2:Contrapositive of P1: If n is odd then 3n + 2 is odd

n=2k+1 where k=0,1,2,...

3n+2=6k+3+2
= 6k+5
=6k+6-1
=2(3k+3) - 1
= even - 1
= odd

QED?

 

[ssd]

3n+2 even
=>3n even
=>n even. This is the basic idea.