I am trying to prove the proposition by contradiction
For all rational $c > 0$, there exists a rational number $x$ such that $x^2 < 2 < (x + c)^2$.
with the negation $x^2 \ge 2 \lor 2 \ge (x+c)^2$. But I can not prove it.
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Let $c > 0$ be rational. Suppose for contradiction that there is no such no negative rational $x$. This means that if $x$ is non-negative and $x^2 < 2$, then we have $(x + c)^2 < 2$. Since $0^2 < 2$, thus $c^2 < 2$, which implies $(2c)^2 < 2$, and by induction we have $(nc)^2 < 2$ for every natural number $n$. We can find an integer $n$ such that $n > 2/c$, which implies that $nc > 2$, which implies that $(nc)^2 > 4 > 2$, contradicting $(nc)^2 < 2$ for all natural numbers $n$.
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You can negate the statement $a < b < c$ to obtain $a < b \implies b \ge c$. Then using induction you can lead a contradiction.
Max
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This is more of a hint or comment than an answer. It might be good to leave it as a comment (which you are not able to do) or label it as a hint so that those reading your answer see that when they rate your answer. – robjohn Oct 09 '14 at 20:27
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If under this assumption $x$ is a no negative number with $x^2<2$, the first alternative is wrong, hence the second must be true and by irrationality of $\sqrt{2}$ we can replace $\ge$ with $>$ in it.