I am working through some properties of $\mathbb{R}$ and I stumbled upon the following theorem:

Theorem 1.21: For every real $x>0$ and every integer $n>0$, there is one and only one positive real $y$ such that $y^n=x$.

The proof of this theorem, as stated in Rudin's Principles of Mathematical Analysis heavily relies on the following inequality: $$ b^n-a^n < (b-a)nb^{n-1}, \ \text{where} \ 0<a<b. $$

I would really like to prove this inequality myself, and I tried rewriting the RHS. This produced $$ b^n-a^n<n(b^n-a\cdot b^{n-1}). $$ Clearly, the value given inside the parentheses on the RHS will be positive, as $b>a$, but I fail to see how this gives me the inequality itself. Does it maybe have to do something with the Archimedean Property of $\mathbb{R}$?

Any help would be much appreciated.

  • 6
    $\begingroup$ Hint: factor $b^n-a^n$ as the product of $b-a$ and something else. $\endgroup$ May 20, 2020 at 13:21
  • 5
    $\begingroup$ $(b^n-a^n)=(b-a)(b^{n-1}+b^{n-2}a+\ldots + a^{n-1})$ $\endgroup$
    – Maryam
    May 20, 2020 at 13:22

1 Answer 1


note that b^n - a^n = (b-a)(b^n-1 + a*b^n-2 + a^2*b^n-3 + ... + b*a^n-2 + a^n-1) ...(1) also since for any k>0 x^k is an increasing function , and hence a < b implies a^k < b^k therefore , a^i * b ^(n-i-1) < or = b^i * b^(n-i-1) = b^n-1 for i = 0,1,2,...,n-1 .... (2). So from (1) and (2) we have b^n - a^n < or =(b-a)nb^n-1 which is what we were required to prove.


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