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I was reading some of the proofs for $\log x < x$ and I noticed a few of them have the proof start off with: "Let $g(x) = x - \log x$". Then they find its derivative to show if the function is increasing or not. I get that part. What I don't understand is how they arrived at choosing "$g(x) = x - \log x$". Is this something that was found that made the proof easier or is there a better reason as to why this function is chosen at the start of the proof?

jameselmore
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David Velasquez
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    A pretty standard way of proving "A< B" is to show that B- A> 0. Here you want to show that log x< x. That gives A= log x and B= x so B- A is x- log(x). – user247327 Sep 04 '15 at 20:53

4 Answers4

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If you want to prove that $f(x) < g(x)$ (in your case, $\log x < x$), this is equivalent to $0 < g(x)-f(x)$ by subtracting $f(x)$ from both sides.

This is a very standard procedure, and is often used.

marty cohen
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$\log x<x$ is the same as $0<x-\log x$, and that's the same as saying the minimum value of $x-\log x$ is more than $0$. Finding minimum values is a topic that the reader is presumed to be accustomed to. The derivative of $x-\log x$ is negative when $x<1$ and positive when $x>1$, so $x\mapsto x-\log x$ decreases as $x$ gets bigger when $x<1$ and increases as $x$ gets bigger when $x>1$. Therefore its minimum occurs when $x=1$. And that minimum is $1-\log1=1$, and that is positive.

Michael Hardy
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Many of the technical results in analysis and calculus deal with the zeroes of a single function, or whether a single function is positive, etc. Basically, they're rewriting the inequality $x>\log x$ into $x-\log x>0$, and then use the available mathematical machinery on that function.

It is easier to compare a slightly complicated function to zero than to compare two simple functions to each other. Both in the technical sense as mentioned, but also because it's conceptually easier to apply derivatives and similar told the correct way.

Arthur
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$\log x<x\iff x-\log x>0$ so it makes sense to show that $g(x)>0$

David Quinn
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