Proving the statement directly: If k3 is even, then k is even.

[zeion]

Homework Statement

a) Let k be any integer. Prove that if k³ is even, then k is even.

Homework Equations

The Attempt at a Solution

Proof by contradiction:

If the hypothesis is true, then k³ cannot be even if k is odd.

Assume k is odd:
k = 2n + 1, such that n is any integer.
k³ = (2n + 1)³
k³ = (2n +1)(2n +1)(2n +1)
k³ = 2(k³ + 5k² + 3k) + 1
Now, (k³ + 5k² + 3k) is any integer.
Therefore k³ is odd if k is odd, hence if k³ is even if k is even.

 

[ravioli]

Looks okay to me, unless you were trying to do a proof by contradiction. Your current proof is one of a contrapositive.

 

[ravioli]

My bad, last statement is iffy. I think it should read if k cubed is even, then k is even

 

[zeion]

Ok thanks

 

[Mark44]

Homework Statement

a) Let k be any integer. Prove that if k³ is even, then k is even.

Homework Equations

The Attempt at a Solution

Proof by contradiction:

If the hypothesis is true, then k³ cannot be even if k is odd.

Assume k is odd:
k = 2n + 1, such that n is any integer.
k³ = (2n + 1)³
k³ = (2n +1)(2n +1)(2n +1)
k³ = 2(k³ + 5k² + 3k) + 1

The line above doesn't make sense to me. If you multiply out the right side, you get 8n³ + 12n² + 6n + 1 = 2(4n³ + 3n + 3) + 1. This is clearly an odd integer, which means that k³ is an odd integer.

Now, (k³ + 5k² + 3k) is any integer.
Therefore k³ is odd if k is odd, hence if k³ is even if k is even.

Another approach is to prove the statement directly.
Assume that k³ is even.
Then k³ = 2m for some integer m.
Because k occurs to the third power, there must be three factors of 2 on the right.
So k³ = 2*2*2*n for some integer n.
Again, because k occurs to the third power, n must have three equal factors, say n = p³.
Hence, k³ = 2*2*2*p*p*p, from which you can easily show that k = 2p, an even integer.