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Given that all rational numbers have the following property: $\displaystyle\left\{ \frac{x}{y} \,\middle|\, x,y\in\mathbb Z\right\}$

it can be inferred therefore that $$\sqrt2=\frac{x}{y}$$ therefore $$2=\frac{x^2}{y^2}$$ and since $2=\frac{2}{1}$, it can be inferred that $y^2=1$, hence $$2=x^2$$ and $$\sqrt2=x$$ hence, by contradiction to $\displaystyle \sqrt2=\frac{x}{y}$, the original statement is false and $\sqrt2$ is in fact an irrational number.

José Carlos Santos
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joshuaheckroodt
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    You have to assume that $x$ and $y$ are coprime to conclude $y^2=1$. – user363120 Sep 25 '17 at 13:22
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    You also haven't stated specifically why $x=\sqrt 2$ is a contradiction - what does it contradict. – Mark Bennet Sep 25 '17 at 13:25
  • Without saying how "it can be inferred that..." the proof is incomplete (and possibly incorrect). It is impossible to judge its correctness without knowing the crucial missing step(s). – Bill Dubuque Sep 25 '17 at 13:36

2 Answers2

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It is wrong. For instance, from the equality $\frac21=\frac{x^2}{y^2}$, you deduce that $2=x^2$ and that $1=y^2$. Why? What about the equality $\frac1{\sqrt2}=\frac{\sqrt2}2$ (which is valid)? Do you deduce from it that $1=\sqrt2$?

José Carlos Santos
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  • Technically it is incomplete not wrong, since we don't know what the OP has in mind for "it can be inferred that..." (which could possibly be correct). – Bill Dubuque Sep 25 '17 at 13:42
  • In the immortal words of Miss Eliza Doolittle: “Not bloody likely!”. Besides, the OP marked my answer as the accepted one. – José Carlos Santos Sep 25 '17 at 13:44
  • In fact it is not too unlikely, as I have seen that occur many times before in arguments like this. And a checkmark on a random website has absolutely nothing to do with mathematical definitions. – Bill Dubuque Sep 25 '17 at 13:45
  • @BillDubuque The sentence "From $\frac{x^2}{y^2}=\frac21$, we can inferr that $2=x^2$ and $1=y^2$" is not an "incomplete" statement, it is a false statement. We cannot infer one from the other. This is not up for debate no matter what OP has in mind. – 5xum Sep 25 '17 at 19:53
  • @5xum You can't possibly judge the OP's (omitted) argument as false without knowing precisely what it is (since there are valid ways to complete it). – Bill Dubuque Sep 25 '17 at 20:12
  • @BillDubuque There is no valid proof that proves that $$\frac{x^2}{y^2} = \frac21 \implies x^2=2, y^2=1$$ is a true statement. – 5xum Sep 25 '17 at 20:37
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    @5xum But that's not what the OP wrote. Rather, the OP's "it can be inferred that" could well mean: wlog the rational $x/y$ is reduced so by Euclid/FTA $,x^2/y^2$ is also reduced so $,x^2/y^2 = 2/1,\Rightarrow,x^2 = 2, y^2 = 1,$ by uniqueness of reduced fractions (Unique Fractionization). That's an argument that many beginners have intuitively in mind, but often don't have the experience to complete to a rigorous proof. – Bill Dubuque Sep 25 '17 at 20:44
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Your proof is incorrect here:

it can be inferred therefore that $$\sqrt2=\frac{x}{y}$$ therefore $$2=\frac{x^2}{y^2}$$ and since $2=\frac{2}{1}$ it can be inferred that $y^2=1$,

This is incorrect. For example, if we set $a=8,b=4$, then $\frac{a}{b}=2$, however we cannot, from $2=\frac 21$,conclude that $b=1$.

Furthermore, you only conclude that $x=\sqrt{2}$, and then claim "this is a contradiction", but it is not clear why this is a contradiction.

5xum
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  • As I remarked in Jose's answer, the proof is incomplete, not incorrect*, since there are valid ways that "it can be inferred that ..." – Bill Dubuque Sep 25 '17 at 20:28