MathLabs

Problem 2

Let a,b,n>1a,b,n>1 be integers and let a,ba,b be bases of two number systems. Using the same digits xn,…,x0x_n,\ldots,x_0, with xn≠0x_n\ne0 and xn−1≠0x_{n-1}\ne0, define An=xnxn−1⋯x0A_n=x_nx_{n-1}\cdots x_0 and An−1=xn−1⋯x0A_{n-1}=x_{n-1}\cdots x_0 in base aa, and Bn=xnxn−1⋯x0B_n=x_nx_{n-1}\cdots x_0 and Bn−1=xn−1⋯x0B_{n-1}=x_{n-1}\cdots x_0 in base bb. Prove that An−1An<Bn−1Bn\frac{A_{n-1}}{A_n}<\frac{B_{n-1}}{B_n} if and only if a>ba>b.
Step 2 of 5: Sum over the digits
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In plain words

One nonzero earlier digit is enough to make the total comparison strict.

anBn>bnAna^nB_n>b^nA_n
Detailed analysis

Multiply the termwise inequalities by xk≥0x_k\ge0 and sum. Since xn−1≠0x_{n-1}\ne0, the k=n−1k=n-1 term is strict. The sums are anBna^nB_n and bnAnb^nA_n.