# property

• Jul 12th 2008, 06:07 PM
particlejohn
property
Fix $\displaystyle b>1$. Prove that $\displaystyle b^{x+y} = b^{x}b^{y}$ for $\displaystyle x,y \in \bold{R}$. We know that $\displaystyle b^{x}b^{y} = \sup B(x) \sup B(y) \geq b^{l}b^{w} = b^{l+w}$ for rational $\displaystyle l \leq x$ and $\displaystyle w \leq y$. Here, $\displaystyle l+w \leq x+y$, where $\displaystyle l+w \in \bold{Q}$. This implies that $\displaystyle \sup B(x) \sup B(y)$ is an upper bound of $\displaystyle B(x+y)$.

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

Note: If $\displaystyle x$ is real, define $\displaystyle B(x)$ to be the set of all numbers $\displaystyle b^t$, where $\displaystyle t$ is rational and $\displaystyle t \leq x$.
• Jul 12th 2008, 08:32 PM
NonCommAlg
Quote:

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

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

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

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

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

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

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

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

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

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

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