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Thread: Spivak's Calculus Question

  1. #1
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    Spivak's Calculus Question

    Prove that if $\displaystyle |x - x_0| < e/2$ and $\displaystyle |y - y_0| < e/2$, then

    $\displaystyle |(x + y) - (x_0 + y_0)| < e$,
    $\displaystyle |(x - y) - (x_0 - y_0)| < e$.

    I'm not looking for a complete solution, just a hint as to where to begin.

    Thanks.
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  2. #2
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    Re: Spivak's Calculus Question

    Quote Originally Posted by RogueDemon View Post
    Prove that if $\displaystyle |x - x_0| < e/2$ and $\displaystyle |y - y_0| < e/2$, then
    $\displaystyle |(x + y) - (x_0 + y_0)| < e$,
    $\displaystyle |(x - y) - (x_0 - y_0)| < e$.

    just a hint as to where to begin.
    $\displaystyle (x+y)-(x_0+y_0)=(x-x_0)+(y-y_0)$
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  3. #3
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    Re: Spivak's Calculus Question

    Will the following identity be of any use in completing the proof?

    $\displaystyle |(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0|$
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    Re: Spivak's Calculus Question

    Quote Originally Posted by RogueDemon View Post
    Will the following identity be of any use in completing the proof?

    $\displaystyle |(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0|$
    That is the only to do it. That is the triangle inequality.
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  5. #5
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    Re: Spivak's Calculus Question

    This is what I've got so far:

    $\displaystyle |(x + y) - (x_0 + y_0 )| < e$

    $\displaystyle |x + y - x_0 - y_0| < e $

    $\displaystyle |x - x_0 + y - y_0| < e $

    $\displaystyle |(x - x_0) + (y - y_0)| < e$

    $\displaystyle |(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0|$


    $\displaystyle |x - x_0| < e/2$

    $\displaystyle 2|x - x_0| < e$

    $\displaystyle |x - x_0| + |x - x_0| < e$

    $\displaystyle |x - x_0| < e - |x - x_0|$

    $\displaystyle |x - x_0| + |y - y_0| < e - |x - x_0| + |y - y_0|$

    $\displaystyle |(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0| < e - |x - x_0| + |y - y_0|$

    $\displaystyle |(x - x_0) + (y - y_0)| < e - |x - x_0| + |y - y_0|$

    Any idea as to how to turn the $\displaystyle e - |x - x_0| + |y - y_0|$ into a simple $\displaystyle e$? Am I even on the right track?
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  6. #6
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    Re: Spivak's Calculus Question

    Given that $\displaystyle |x-x_0|<\frac{e}{2}~\&~|y-y_0|<\frac{e}{2}$ then
    $\displaystyle |(x-x_0)+(y-y_0|\le |(x-x_0)|+|(y-y_0|<\frac{e}{2}+\frac{e}{2}=e.$
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