Solving problems.

**felixmcgrady** · September 8th 2009, 02:39 AM

Porve that b^(r+s) = b^r * b^s, r and s are rationals.

**red_dog** · September 9th 2009, 02:04 AM

Let $r=\frac{m}{n}, \ s=\frac{p}{q}, \ m,n,p,q\in\mathbb{N}, \ n,q\neq 0$

Then $b^{r+s}=b^{\displaystyle\frac{m}{n}+\frac{p}{q}}=b ^{\displaystyle\frac{mq+np}{nq}}=$

$=\displaystyle\sqrt[nq]{b^{mq+np}}=\sqrt[nq]{b^{mq}b^{np}}=$

$=\sqrt[nq]{b^{mq}}\cdot\sqrt[nq]{b^{np}}=b^{\frac{mq}{nq}}\cdot b^{\frac{np}{nq}}=b^{\frac{m}{n}}\cdot b^{\frac{p}{q}}=b^r\cdot b^s$

In a similar way you can prove if r or s or both are negative.

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