# exponents: proofs

• Jun 14th 2007, 01:49 PM
tukeywilliams
exponents: proofs
Prove (i) $\displaystyle x^{n}y^{n} = (xy)^{n}$ for nonzero real numbers $\displaystyle x,y$ and integers $\displaystyle n$

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

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

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

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

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

Thanks
• Jun 15th 2007, 04:04 AM
topsquark
They look good to me.

-Dan