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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 $\frac{17}{10}$.) –  alex.jordan Apr 30 '12 at 16:25
    
@DavidMitra Why are you cubing? –  Thomas Andrews Apr 30 '12 at 16:26
    
@ThomasAndrews erm, just a bit off at the moment :) –  David Mitra Apr 30 '12 at 16:26
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Note that 153 is divisible by 9. Clear fractions and square both sides. –  Mark Bennet Apr 30 '12 at 16:27
    
Oh my, for what a stupidity I asked.. My self-punishment is posted below, thank you all and sorry :) –  Straightfw Apr 30 '12 at 16:40

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up vote 10 down vote accepted

$\sqrt3>\frac{153}{90}$ 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! –  Straightfw Apr 30 '12 at 16:39

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