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  1. #1
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    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 .
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    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.
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