# property

• Jul 12th 2008, 07:07 PM
particlejohn
property
Fix $b>1$. Prove that $b^{x+y} = b^{x}b^{y}$ for $x,y \in \bold{R}$. We know that $b^{x}b^{y} = \sup B(x) \sup B(y) \geq b^{l}b^{w} = b^{l+w}$ for rational $l \leq x$ and $w \leq y$. Here, $l+w \leq x+y$, where $l+w \in \bold{Q}$. This implies that $\sup B(x) \sup B(y)$ is an upper bound of $B(x+y)$.

Is this correct? We want to prove this property for all reals.

Note: If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t \leq x$.
• Jul 12th 2008, 09:32 PM
NonCommAlg
Quote:

Originally Posted by particlejohn
Fix $b>1$. Prove that $b^{x+y} = b^{x}b^{y}$ for $x,y \in \bold{R}$. We know that $b^{x}b^{y} = \sup B(x) \sup B(y) \geq b^{l}b^{w} = b^{l+w}$ for rational $l \leq x$ and $w \leq y$. Here, $l+w \leq x+y$, where $l+w \in \bold{Q}$. This implies that $\sup B(x) \sup B(y)$ is an upper bound of $B(x+y)$.

Is this correct? We want to prove this property for all reals.

Note: If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t \leq x$.

$(*)$ the limit of the sequence $\{ b^{\frac{1}{n}}\}$ is 1 as $n \rightarrow \infty.$ this is because since b > 1, we have $b^{\frac{1}{n}} = 1 + c_n,$ for some $c_n > 0.$

thus: $b=(1+c_n)^n \geq 1 + nc_n.$ hence $0< c_n \leq \frac{b-1}{n}.$ therefore $\lim_{n\to\infty} c_n =0,$ i.e. $\lim_{n\to\infty}b^{\frac{1}{n}}=1.$

back to your question: first let $t, s \in \mathbb{Q}$ such that $t \leq x, \ s \leq y.$ then $t+s \leq x+y.$ thus by definition: $b^t b ^s = b^{t+s} \leq b^{x+y}.$

now fix $s$ and take $\sup$ of both sides over all rational numbers $t \leq x$ to get $b^xb^s \leq b^{x+y}.$ now take $\sup$ of both sides of this

inequality over all rational numbers $s \leq y,$ to get $b^xb^y \leq b^{x+y}. \ \ \ (1)$

next let $n \in \mathbb{N}$ and choose $t,s \in \mathbb{Q}$ such that $t \leq x, \ s \leq y$ and $x - t \leq \frac{1}{2n}, \ y-s \leq \frac{1}{2n}.$ let $r \in \mathbb{Q}$ such that $r \leq x+y.$

then: $r \leq x+y \leq t+s + \frac{1}{n}.$ so: $b^r \leq b^t b^s b^{\frac{1}{n}} \leq b^x b^y b^{\frac{1}{n}}.$ taking $\sup$ of both sides over all rational numbers $r \leq x+y$ gives

us: $b^{x+y} \leq b^x b^y b^{\frac{1}{n}}.$ now let $n \rightarrow \infty.$ then using $(*)$ we'll get: $\ b^{x+y} \leq b^xb^y,$ which with $(1)$ complete the proof. Q.E.D.

Note: throughout the proof i've assumed that we know all basic properties for rational powers of a real number.