Question

6. Fix b (a) If m, n, p, q are integers, n > 0, q > 0, and r = mln-plg, prove that Hence it makes sense to define y (b")1/n. (b) Prove that b… = b,b" if r and s are rational. (c) If x is real, define B(x) to be the set of all numbers b', where t is rational and tSx. Prove that b-sup B(r) ris rational. Hence it b" = sup B(x) for every realx (d) Prove thatbb' for all real x and y.

Answer 1

SOLUTION:

It is left as an exercise ti show that (b^m)^q = b^m q for any two integers m, q and real b > 0 using the associativity of multiplication (and induction) Using this fact [(b^m)^1/n]^n q = {[(b^m)^1/n}^q by associativity of multiplication = {b^m)^q by definition of the n- t h root = b^m q by associativity of multiplication Likewise [(b^p)^1/q]^n q = {[(b^p)^1/q]^q}^n by associativity of multiplication = {b^p}^n by definition of the q- t h root = b^n q by associativity of multiplication Consequently [(b^m)^1/n]^n q = b^m q = b^n q = [(b^p)^1/q]^n q where the middle equality follows from m q = p n (since m/n = p/q Taking am m q -t h root on both sides of the equation gives the result (b^m)^1/n = (b^p)^1/q b) Write e = m/n and s = p/q then r + s = m q + n p/n q, so that by the definition gives in part (a)