1.Assume $a$ and $b$ are irrational numbers. Find a counterexample showing that it is possible $a-b$ is rational.
- 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.
- 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.