Prove directly and indirectly $\displaystyle x<1$ then $\displaystyle x^2=>1$
for direct proof I arrived at
$\displaystyle x^2 = \frac{r-1}{r}$
For indirectly I need some help.. a nudge in the right direction.
If $\displaystyle x < 1$, then $\displaystyle x^2 = x \times x = (-1 \times -x) \times (-1 \times -x) = -x \times -x$. Since $\displaystyle x < 1$, $\displaystyle -x > 1$, and thus $\displaystyle -x \times -x > 1$.
By the way, your statement is false. If $\displaystyle x < 1$, then $\displaystyle x^2 > 1$, and $\displaystyle x^2 \neq 1$.
It is not "false" but it is unnecessary to write that $\displaystyle x^2 \bold{=}> 1$ since $\displaystyle x^2 \neq 1$. In some situations, adding a value for a variable can be viewed as an error. I was just pointing that out.Prove directly and indirectly $\displaystyle x < 1$ then $\displaystyle x^2 => 1$
Let's see if we can make some order here: do you want to prove that $\displaystyle x<1\Longrightarrow x^2\geq 1$ ?? This, of course, is false: just take $\displaystyle x=\frac{1}{2}\,,\,\,x^2=\frac{1}{4}<1$ ...
But this is trivial, so I wonder what did you really mean...
Tonio
$\displaystyle x<-1\ \Rightarrow\ x^2>1$
Since if $\displaystyle 0<x<1\ \Rightarrow\ x^2<1$
$\displaystyle x=\frac{a}{b},\ a<b,\ x^2=\frac{a^2}{b^2}$
$\displaystyle a<b\ \Rightarrow\ a^2<b^2\ if\ a,b>0\ \Rightarrow\ \frac{a^2}{b^2}<1$
However, if $\displaystyle x<-1,\ x=-\frac{b}{a},\ a<b$