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

  1. #1
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Saviour View Post
    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.
    Hmm...are you sure about this.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Hmm...are you sure about this.
    lol no
    Last edited by Saviour; Feb 23rd 2010 at 10:22 PM.
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  4. #4
    Super Member Bacterius's Avatar
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    A square is always positive.
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    Quote Originally Posted by Bacterius View Post
    A square is always positive.
    I'm so lost...

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

    let k<r
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  6. #6
    Super Member Bacterius's Avatar
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    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 $\displaystyle x^2 \neq 1$.
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  7. #7
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    Quote Originally Posted by Bacterius View Post
    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 $\displaystyle x^2 < 1$
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  8. #8
    Super Member Bacterius's Avatar
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    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.
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  9. #9
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    Quote Originally Posted by Saviour View Post
    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
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  10. #10
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    $\displaystyle x<-1\ \Rightarrow\ x^2>1$

    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 $\displaystyle x<-1,\ x=-\frac{b}{a},\ a<b$
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  11. #11
    Super Member Bacterius's Avatar
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    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
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  12. #12
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    I reckon Saviour was supposed to start with x<-1,
    otherwise he'd have to use 2 cases,
    just pointing out an option, not slipping up.
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  13. #13
    Super Member Bacterius's Avatar
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    That's what I thought too, it seemed such a typical question that I just flew over it, now hopefully he made a typo
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  14. #14
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    Quote Originally Posted by Bacterius View Post
    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
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