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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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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
  • 225,327