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Let $R$ be a relation on the set $\Bbb R$ of real numbers where real numbers $x,y$ satisfy $xRy$ if and only if $e^{x-y}$ is an integer. Is $R$ an equivalence relation on $\Bbb R$? Is it a partial order?

I have proved it's reflexive let $x=1$. for reflexivity $xRx$.

so $e^{x-x}= e^{1-1}=e^0 =1$. This also satisfies the result being an integer. Therefore the relation is reflexive.

To check is the relation is symmetric it means. $xRy$ then $yRx$. The problem I am facing is I can't seem to find an $x$ and $y$ that satisfies $e^{x-y}$ being an integer.

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Maybe you mean $e^{x-y}$ is an integer, maybe you mean $e^x-y$ is an integer. which is it? –  André Nicolas Dec 30 '12 at 18:40
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I think it is about time you try to enhance seriously the way you write your posts: use LaTeX for mathematics (in the FAQ section you canfind some directions), use questions mark, periods where required, etc. –  DonAntonio Dec 30 '12 at 18:42
    
Based on the partial solution I guess that it is $e^{x-y}$. –  Sigur Dec 30 '12 at 18:42
    
Do you really think that to shwo reflexivity it suffices to consider merely the case $x=1$? –  Hagen von Eitzen Dec 30 '12 at 18:43
    
@HagenvonEitzen Thats my own understanding.What way would you have done it? –  Jack welch Dec 30 '12 at 18:44

2 Answers 2

You relation is:

  • reflexive: $\forall x \in \mathbb{R}, e^{x-x} = e^0 = 1$ so $xRx$

  • transitive: $\forall x, y, z \in \mathbb{R}$, if you have $xRy$ and $yRz$, you have $e^{x-y}\in \mathbb{N}$ and $e^{y-z}\in \mathbb{N}$ so $e^{x-z} = e^{x-y+y-z} = e^{x-y}e^{y-z}\in \mathbb{N}$ so you also have $xRz$

  • not symmetric: $e^{\ln(2)-0} = e^{\ln(2)}=2\in \mathbb{N}$ so $\ln(2)R0$ but $e^{0-\ln(2)} = e^{-\ln(2)} = \cfrac{1}{e^{\ln(2)}} = \cfrac{1}{2} \not\in \mathbb{N}$ so you don't have $0 R \ln(2)$

  • antisymmetric: $\forall x, y \in \mathbb{R}$, if you have both $xRy$ and $yRx$, it means that $e^{x-y}\in\mathbb{N}$ and $\cfrac{1}{e^{x-y}}\in\mathbb{N}$. If $e^{x-y}>1$, then $0<\cfrac{1}{e^{x-y}}<1$ so $\cfrac{1}{e^{x-y}}\not\in\mathbb{N}$ so $e^{x-y}=1$ so $x-y=0$ ie $x=y$

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If someone knows how to render \not R $\not{R}$ properly, please edit. –  xavierm02 Dec 30 '12 at 18:55
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(+1) although it is "symmetric". –  TMM Dec 30 '12 at 19:10

I will assume that "$e^{x-y}$ is an integer" is intended. Reflexivity is obvious: $e^{x-x}=e^0=1$ for all $x$.

About symmetry, note that $e^{y-x}=\dfrac{1}{e^{x-y}}$. So presumably most of the time, when $e^{x-y}$ is an integer, $e^{y-x}$ is not an integer. In fact, if $e^{x-y}$ is an integer, then $e^{y-x}$ is also an integer precisely if $x=y$. You will find this useful in dealing with the question about partial order.

For an explicit example, let $x=\ln 2$ and $y=0$. Then $e^{x-y}=e^{\ln 2}=2$, an integer, while $e^{y-x}=e^{-\ln 2}=\dfrac{1}{2}$ is not an integer.

Now to deal with the question about order, suppose that $e^{x-y}$ is an integer, and $e^{y-z}$ is an integer. Can we conclude that $e^{x-z}$ is an integer?

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..How do you pick explicit examples like log2 –  Jack welch Dec 30 '12 at 19:05
    
We want to find $u$ such that $e^u$ is an integer. Recall that $e^{\ln t}=t$ for all (positive) $t$. So $\ln 2$ is a good choice, or $\ln(17)$. –  André Nicolas Dec 30 '12 at 19:08
    
@ Andre Thank you –  Jack welch Dec 30 '12 at 19:14

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