direct and indirect proof

**murielx** · February 26th 2010, 07:55 PM

hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

I need to prove directly and by contradiction that

if x is a positive real number and x < 1, then x^2 < 1

all help will be greatly appreciated

**Prove It** · February 26th 2010, 08:14 PM

Originally Posted by murielx

hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

I need to prove directly and by contradiction that

if x is a positive real number and x < 1, then x^2 < 1

all help will be greatly appreciated

If $0 < x < 1$ , then we can write $x = \frac{1}{y}$ , where $y > 1$ .

So $x^2 = \left(\frac{1}{y}\right)^2 = \frac{1}{y^2}$ .

Since $y > 1, y^2 > 1$ .

Therefore $\frac{1}{y^2} < 1$ .

Therefore $x^2 < 1$ .

**Prove It** · February 26th 2010, 08:16 PM

Originally Posted by murielx

hello all, someone posted a similar question to this but he wrote the question wrong so the subsequent answers were wrong.

I need to prove directly and by contradiction that

if x is a positive real number and x < 1, then x^2 < 1

all help will be greatly appreciated

$0 < x < 1$ .

Assume $x^2 \geq 1$

Then $\sqrt{x^2} \geq \sqrt{1}$

$|x| \geq 1$

$x \leq -1$ or $x \geq 1$ .

This contradicts what we said originally. So $x^2 < 1$ .

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