# direct and indirect proof

• Feb 26th 2010, 07:55 PM
murielx
direct and indirect proof
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
• Feb 26th 2010, 08:14 PM
Prove It
Quote:

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

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

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

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

Therefore $\displaystyle x^2 < 1$.
• Feb 26th 2010, 08:16 PM
Prove It
Quote:

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

$\displaystyle 0 < x < 1$.

Assume $\displaystyle x^2 \geq 1$

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

$\displaystyle |x| \geq 1$

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

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