Discrete Math Questions? (checking my answers) 1.Assume $a$ and $b$ are irrational numbers. Find a counterexample showing that it is possible $a-b$ is rational. 


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*My answer: Suppose $a$ and $b$ are equal to $\sqrt{2}$.  Therefore, $a-b= \sqrt{2}- \sqrt{2}=0$.  This proof by counterexample that $a$ and $b$ are irrational but $a-b$ is rational?...(is this correct)?


2.If $x+1$ is odd, prove that $(x+1)^3$ is odd.


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*My answer: let $x+1$ be odd therefore $x+1=2k+1$ for some integer $k$.
Suppose $x+1$ is not odd. Then $x=2k$ where 
\begin{align}
(x+1)^3 & = x^3+3x^2+3x+1\\  
        & = (2k)^3+3(2k)^2+3(2k)+1\\
        & = 2(4k^2+6k^2+3k)+1\\
        & = 2(a)+1 
\end{align}
for some integer $a$ where $a=4k^2+6k^2+3k$.  Therefore, this proves that $(x+1)^3=2a+1$ is odd.


1. Done.
2. Suppose $x+1$ is odd. Then $x+1=2k+1$ for some integer $k$. Then
$$
\begin{aligned}[t]
(x+1)^3 
= (2k+1)^3 
&= (2k)^3+3(2k)^2+3(2k)+1\\ 
&= 8k^3+12k^2+6k+1\\
&= 2(4k^3+6k^2+3k)+1.
\end{aligned}
$$
Thus, $(x+1)^3=2m+1$, where $m$ is the integer $4k^3+6k^2+3k$, so $(x+1)^3$ is odd. Therefore, we have shown that if $x+1$ is odd, then $(x+1)^3$ is odd.
