Let $x$ be a real number. Prove that if $x^2\lt x$, then $x\lt1$

Question

I do not know how to proceed. I assume its proof is by contrapositive?

user id:307825 10k · Accepted Answer

$$x^2\lt x$$ $$x^2-x\lt 0$$ $$x(x-1)\lt 0$$ Thus, either $x\lt 0$ and $x\gt 1$

or

$x\gt0$ and $x\lt1$.

Clearly, the former is not possible.

Hence, $$0\lt x\lt1$$

user id:228959 3k · Accepted Answer

3

Proof by contrapositive is the easiest: let $x \ge 1$, then $x^2 = x \cdot x >= x$. Q.E.D.

asked 2017-02-06

user id:228959

3k

0

You don't exclude the possibility of $x$ being negative. – 2017-02-06
0

I'm *assuming* $x$ is positive to begin with. That's the whole idea of proving the contrapositive. – 2017-02-06
0

Ah, sorry, I have misread the question - based on the top answer I thought you are supposed to prove $0 – 2017-02-06

user id:589 170k · Accepted Answer

$x^2\lt x$ implies $x>0$ since $x^2\ge0$ for all $x$.

Therefore we can cancel $x$ in $x^2\lt x$ and get $x < 1$.

user id:401635 9k · Accepted Answer

1

$x^2 < x$

$\frac{x^2}{x} < 1$

$x < 1$

asked 2017-02-06

user id:401635

9k

0

You need $x>0$ for the second step. – 2017-02-06
0

@lhf but how? If I answer it like $x^2 - x<0$ then I know. But in this case don't know. – 2017-02-06
0

@lhf as $x>x^2\geq 0\implies x>0$ – 2017-02-08

user id:406412 1k · Accepted Answer

exactly. if $x\geq 1$ (1), as $x>0$ and $x=x$ (2), (1)$\dot{}$(2) gives you $x^2\geq x$

5 Answers 5