# Thread: Can anyone help me with 2 homework questions?

1. ## Can anyone help me with 2 homework questions?

Been stuck on these two for a while now. Would anyone be able to help me with them? Thanks!

#1.) Prove by contradiction that for all n ∈ Z, if n^3 + 5 is odd, then n is even

#2.) Prove using a direct proof that for every rational number x, if x
0, then 1/x is rational

2. ## Re: Can anyone help me with 2 homework questions?

1)
Assume n is not even. Set n=2k+1 where k is a natural number
Show that the assumption is false because n^3 +5 cannot be even when n=2k+1.

2)
Let $x=\frac{a}{b}$
Show that 1/x can be written as a quotient.

3. ## Re: Can anyone help me with 2 homework questions?

Originally Posted by Shakarri
1)
Assume n is not even. Set n=2k+1 where k is a natural number
Show that the assumption is false because n^3 +5 cannot be even when n=2k+1.

2)
Let $x=\frac{a}{b}$
Show that 1/x can be written as a quotient.
Thank you!!

4. ## Re: Can anyone help me with 2 homework questions?

Originally Posted by mgk501
Been stuck on these two for a while now. Would anyone be able to help me with them? Thanks!

#1.) Prove by contradiction that for all n ∈ Z, if n^3 + 5 is odd, then n is even

#2.) Prove using a direct proof that for every rational number x, if x
0, then 1/x is rational
It's easier to prove the first question using the contrapositive. Your statement is equivalent to saying if n is odd, then $\displaystyle n^3 + 5$ is even. If n is odd, we can write it as n = 2m + 1. So

\displaystyle \begin{align*} n^3 + 5 &= \left( 2m + 1 \right) ^3 + 5 \\ &= \left( 2m \right) ^3 + 3 \left( 2m \right) ^2 + 3 \left( 2m \right) + 1 + 5 \\ &= 8m^3 + 12m^2 + 6m + 6 \\ &= 2 \left( 4m^3 + 6m^2 + 3m + 3 \right) \end{align*}

which is clearly even. Q.E.D.

5. ## Re: Can anyone help me with 2 homework questions?

"Proving the contrapositive" is a proper subset of "proof by contradiction".

6. ## Re: Can anyone help me with 2 homework questions?

Originally Posted by HallsofIvy
"Proving the contrapositive" is a proper subset of "proof by contradiction".
I don't see how that could possibly be. In my example I did NOT reach a contradiction...