1. ## exponents: proofs

Prove (i) $x^{n}y^{n} = (xy)^{n}$ for nonzero real numbers $x,y$ and integers $n$

Prove (ii) $x^{m+n} = x^{m}x^{n}$ for nonzero real numbers $x$ and integers $m,n$.

Use the definition: $x^{-m} = \frac{1}{x^{m}}$ for $m > 0$.

23. (i) Proof: We use induction on $n$ . Case #1: $n > 0$. Base case: For $n = 0$, $x^{0}y^{0} = 1 = (xy)^{0}$. Inductive step: Suppose now as inductive hypothesis that $x^{k}y^{k} = (xy)^{k}$ for any real numbers $x, y$ and for a non-negative integers $n$. Then $xy(x^{k}y^{k}) = x^{k+1}y^{k+1} = xy(xy)^{k} = (xy)^{k+1}$. Case #2: $n < 0$. Base case: For $n = 0$, $\frac{1}{x^{0}}\frac{1}{y^{0}} = \frac{1}{(xy)^{0}} = 1$ for $n > 0$. Suppose now as induction hypothesis that $\frac{1}{x^{k}}\frac{1}{y^{k}} = \frac{1}{(xy)^{k}}$ for some positive integer $k$. Then $\frac{1}{xy}\frac{1}{x^{k}}\frac{1}{y^{k}} = \frac{1}{x^{k+1}}\frac{1}{y^{k+1}} = \frac{1}{xy}\frac{1}{(xy)^{k}} = \frac{1}{(xy)^{k+1}}$. Conclusion: Hence, by induction, $x^{n}y^{n} = (xy)^{n}$ for any real numbers $x, y$ and for integers $n$. $\square$

(ii) Proof: We use induction on $n$. Case #1: $m,n > 0$. Base case: For $n = 0$, $x^{m+0} = x^{m} = x^{m}x^{0}$. Inductive step: Suppose now as inductive hypothesis that $x^{m+k} = x^{m}x^{k}$ for any real numbers $x,y$ and for non-negative integers $m,n$. Then $x \cdot x^{m+k} = x^{m+k+1} = x\cdot x^{m}x^{k} = x^{m}x^{k+1}$. Case #2: $m,n < 0$. Base case: For $n = 0$, $\frac{1}{x^{n}} = \frac{1}{x^{n}}$. Inductive step: Suppose now as inductive hypothesis that $\frac{1}{x^{m+k}} = \frac{1}{x^{m}x^{k}}$ for some positive integer $k$. Then $\frac{1}{x}\frac{1}{x^{m+k}} = \frac{1}{x^{m+k+1}} = \frac{1}{x}\frac{1}{x^{m}x^{k}} = \frac{1}{x^{m}x^{k+1}}$. Case #3: $m >0, n< 0$. Base case: For $n = 0$, $x^m = x^m$. (a) $x^{m+n} > 0$ or (b) $x^{m+n} < 0$. Inductive step: If (a), then suppose as inductive hypothesis that $x^{m+k} = \frac{x^{m}}{x^{k}}$. Then $x \cdot x^{m+k} = x^{m+k+1} = x \cdot \frac{x^{m}}{x^{k}} = \frac{x^{m+1}}{x^{k}} = \frac{x^{m}}{x^{k+1}}$. Conclusion: Hence, by induction, $x^{m+n} = x^{m}x^{n}$ for any real numbers $x$ and for $m,n$. $\square$

For (ii) am I considering the cases correctly? Am I doing this incorrectly?

Thanks

2. They look good to me.

-Dan