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For the real numbers $x=0.9999999\dots$ and $y=1.0000000\dots$ it is the case that $x^2<y^2$. Is it true or false? Prove if you think it's true and give a counterexample if you think it's false.

Ben Grossmann
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Siamoka
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  • Are there dots after the digits '9' or aren't there? – Berci Jun 14 '13 at 00:40
  • yes there are, the dot represent that it will going forever – Siamoka Jun 14 '13 at 00:42
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    Solved [here](http://math.stackexchange.com/q/287311/73324) but also [here](http://math.stackexchange.com/q/281492/73324), depending on your mathematical sophistication. – vadim123 Jun 14 '13 at 00:44
  • How so counterexample? There is absolutely nothing to choose in this question, so it is just a true or false question (in fact false), no example is involved. – Marc van Leeuwen Jun 14 '13 at 09:44

1 Answers1

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Since $x=y$ (that is, $y = 0.\overline{9}=\sum_{n=1}^\infty \frac9{10^n}=1=x$), it must be the case that $x^2=y^2$. Thus, your statement is false.

Ben Grossmann
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