Consider the following \"proof\": Suppose a and b are non-zero real numbers with

Question

Consider the following "proof":

Suppose a and b are non-zero real numbers with a<b. Taking reciprocals is applying the function f(x)=1/x to both sides of the inequality. Since f'(x)= -1/x^2 <0 for all non-zero real numbers, f(x) must be strictly decreasing, from which we conclude 1/a > 1/b.

However, this can't be true, since a=-1, b=1 is a counterexample.

(a). What's wrong with the proof?

(b). Under the extra conditions on a and b does the proof actually work?

(c). State and prove a result for any cases not covered by (b).

Explanation / Answer

(a) Proof can be true for f'(x) but not for f(x) as former has a term of x2 which will nullify the effect of sign whether it is plus or minus but f(x)= 1/x which will not be true for all the reall numbers.

It will be true for positive real numbers only.

(b) Yes, proof will work under extra condition that both a and b are positive real numbers with a<b.

(c) We can consider any case by taking value of a and b as integers or decimal numbers but they have to be positive numbers.

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