Thread: 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 $.

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.