Thread: Prove x< 1 then x^2 =>1

1. Prove x< 1 then x^2 =>1

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.

2. Originally Posted by Saviour
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.

3. Originally Posted by Drexel28
lol no

4. A square is always positive.

5. Originally Posted by Bacterius
A square is always positive.
I'm so lost...

$\displaystyle x^2 = \frac{r-k}{r}$

let k<r

6. 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 $\displaystyle x^2 \neq 1$.

7. Originally Posted by Bacterius
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 $\displaystyle x^2 \neq 1$.
Bloody hell I wrote the wrong symbol it's ment to be $\displaystyle x^2 < 1$

8. Prove directly and indirectly $\displaystyle x < 1$ then $\displaystyle x^2 => 1$
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.

9. Originally Posted by Saviour
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.

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

10. $\displaystyle x<-1\ \Rightarrow\ x^2>1$

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 $\displaystyle x<-1,\ x=-\frac{b}{a},\ a<b$

11. OMG enormous mistake from me, I thought from the start you had written $\displaystyle x < -1$. My mistake ! Then yes the statement is clearly wrong. It seemed so obvious, I must've been confused

EDIT : haha Archie Meade fell in the same booby trap

otherwise he'd have to use 2 cases,
just pointing out an option, not slipping up.

13. That's what I thought too, it seemed such a typical question that I just flew over it, now hopefully he made a typo

14. Originally Posted by Bacterius
That's what I thought too, it seemed such a typical question that I just flew over it, now hopefully he made a typo

Pretty confusing stuff: what did the OP really mean?
I hereby propose to let the OP to correct his/her own post if he wants some attention: it's the minimum one must require from someone asking for help, namely to post CORRECTLY the question.

Tonio