1. ## b^(x+y)=b^x*b^y

let b>1. My textbook has asked me to do a prelimanary proof which I have answered

Define B(x) to be the set with elements b^t where t E Q and t<_ x, where x is real.

Show Sup(B(r))=b^r where r E Q

r>_t so b^r >_ b^t

Since Q is dense in R, for all y < r, there is a c E Q such that y<c<r, implying b^y<b^c<b^r so b^y is not an upperbound. Note all reals > 1 can be expressed b^y for some real y.

Hence b^x=Sup(B(x)) for real x.

So now to the proof in the title. b^x*b^y =Sup(B(x))Sup(B(y)) for real x,y

B(x+y)={b^t:t E Q, t<x+y}. I see I have to show Sup(B(x))Sup(B(y))=Sup(B(x+y))
but can't see how to proceed. Thanks

2. ## Re: b^(x+y)=b^x*b^y

Originally Posted by Duke
let b>1. My textbook has asked me to do a prelimanary proof which I have answered

Define B(x) to be the set with elements b^t where t E Q and t<_ x, where x is real.

Show Sup(B(r))=b^r where r E Q

r>_t so b^r >_ b^t

Since Q is dense in R, for all y < r, there is a c E Q such that y<c<r, implying b^y<b^c<b^r so b^y is not an upperbound. Note all reals > 1 can be expressed b^y for some real y.

Hence b^x=Sup(B(x)) for real x.

So now to the proof in the title. b^x*b^y =Sup(B(x))Sup(B(y)) for real x,y

B(x+y)={b^t:t E Q, t<x+y}. I see I have to show Sup(B(x))Sup(B(y))=Sup(B(x+y))
but can't see how to proceed. Thanks
The idea here is that the textbook defines b^x (for a real number x) to be $b^x \overset{\text{d{e}f}}= \sup(B(x)).$ The point of the preliminary proof is to check that this coincides with the existing definition of b^x in the case where x is rational.

So now you have two real numbers x and y, and you want to show that $\sup(B(x+y)) = \sup(B(x))\sup(B(y)).$ Notice first that if r, s are rational numbers with r<x and s<y then r+s<x+y. You also know that $b^rb^s = b^{r+s} \leqslant \sup(B(x+y))$ (because the index laws are known to hold for rational powers). Taking the sup over all such r and s, you deduce that $\sup(B(x))\sup(B(y)) \leqslant B(x+y)$.

To get the reverse inequality, you need to show that if t is any rational number less than x+y, then it is possible to express t in the form t=r+s, with r<x and s<y. It then follows that $b^t = b^rb^s \leqslant \sup(B(x))\sup(B(y))$. Taking the sup over t, you get $\sup(B(x+y)) \leqslant \sup(B(x))\sup(B(y)).$

Notice that if t < x+y then t – y < x. Therefore there is a rational number r with $t - y < r < x.$ If you define $s = t-r$ then s<y. That justifies the assertion at the start of the previous paragraph (about expressing t in the form r+s).

3. ## Re: b^(x+y)=b^x*b^y

thanks. I did get half of that myself. I think you know the book this is from. Do you think it is difficult? Anyway in another part he states that A and B have a one to one correspondance iff they have the same number of elements. Surely the if part is incorrect.