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How do I prove that, for example, $\sqrt{3}>\frac{153}{90}$? I can't represent it in any other way than periodic fraction and showing by "Hey, look at the calculator, it's bigger!" doesn't look like a good idea :) Representing as a difference of other powers also doesn't seem to work.. How can I elegantly prove this?

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    Are you allowing yourself to check that $3\cdot90^2>153^2$? (Also, $\frac{153}{90}$ reduces to $\frac{17}{10}$.)2012-04-30
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    @DavidMitra Why are you cubing?2012-04-30
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    @ThomasAndrews erm, just a bit off at the moment :)2012-04-30
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    Note that 153 is divisible by 9. Clear fractions and square both sides.2012-04-30
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    Oh my, for what a stupidity I asked.. My self-punishment is posted below, thank you all and sorry :)2012-04-30

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$\sqrt3>\frac{153}{90}$ if and only if $3>\left(\frac{153}{90}\right)^2=\frac{23409}{8100}$. Since $3=\frac{24300}{8100}>\frac{23409}{8100}$, the inequality is clearly true.

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    OMG, thank you very much. Some say there aren't stupid question but mine proves the contrary.. :D Sorry and thank you again!2012-04-30