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I was just thinking about how to prove $\sqrt{5} \leq 3$. I believe I can prove by contradiction by saying suppose $\sqrt{5} > 3$, then it must be true also that $5 > 9$ by squaring both sides, but this is absurd so it must be that $\sqrt{5} \leq 3$. However, I was thinking that this doesn't seem valid since we can suppose $\sqrt{5} > -3$ then squaring both sides gives $5 > 9$, which is also absurd! By we know $\sqrt{5} > -3$, so is something I'm doing not valid? I know this is a simple thing, but I was just trying to prove an irrational number is less than a certain rational number and am stumped.

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    $\begingroup$ If $a>b>0$ then $a^2>b^2$. It is not true if $b<0$. $\endgroup$ – Thomas Andrews Feb 25 '17 at 0:43
  • $\begingroup$ Consider a right triangle with legs $1$ and $2$. Then $\sqrt{5}=c\leq a+b=3$. $\endgroup$ – Christian Blatter Feb 25 '17 at 10:36
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Squaring both sides of an inequality does not preserve the inequality unless both sides were originally positive.

The long way: suppose $\sqrt{5} > 3$. Multiplying both sides by $3$, we have $3\sqrt{5} > 9$. Since $3 < \sqrt{5}$, $3\sqrt{5} < \sqrt{5}^2 = 5$. So $5 > 3\sqrt{5} > 9$, a contradiction.

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    $\begingroup$ "Since $3<\sqrt{5}$"? Sorry, that can't be right, unless you change it into "Since we supposed $3<\sqrt{5}$..... $\endgroup$ – imranfat Feb 25 '17 at 2:58
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    $\begingroup$ @imranfat the answerer is assuming that $\sqrt{5}>3$. $\endgroup$ – Xam Feb 25 '17 at 3:01
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    $\begingroup$ @Xam, yes, it just looks odd... $\endgroup$ – imranfat Feb 25 '17 at 3:03
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Here is another way to look at this problem.

The square-root function, $f(x)=\sqrt x$, is increasing on its domain because $f'(x)>0$ on $(0, +\infty)$. Since $3=\sqrt 9$,

$3=f(9)$.

And $\sqrt 5=f(5)$.

But since $f(x)=\sqrt x$ is increasing on its domain, by the definition of increasing (and because $5<9$):

$f(5) <f(9)$.

Substitute the meanings of $f(5)$ and $f(9)$ and get:

$\sqrt 5 <3$.

Q.E.D.

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  • $\begingroup$ Thank you for this, unnecessary contradiction proofs are so ugly. $\endgroup$ – mathematician Feb 25 '17 at 3:26
  • $\begingroup$ How would I prove the square-root function is increasing? I don't like making assumptions like that. $\endgroup$ – DrBistro240 Feb 25 '17 at 14:44
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    $\begingroup$ @DrBistro240 The derivative of $\sqrt x$ is always greater than 0, so $\sqrt x$ is increasing on its domain. You could check it with computations. (Post edited) $\endgroup$ – Timothy Cho Feb 26 '17 at 19:10
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Hint: $\;\;3-\sqrt{5} \,=\, \cfrac{4}{3+\sqrt{5}}$

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Squaring is not an equivalence transformation.

For example $x=3$ squared becomes $x^2=9$ which has solutions $\pm 3$ rather than $3$ alone.

But you can argue as follows

If you assume $\sqrt{5}>3$ , you can conclude $5=\sqrt{5}\cdot \sqrt{5}>\sqrt{5}\cdot 3>3\cdot 3=9$

because it is clear that $\sqrt{5}$ and $3$ are positive.

This way you get the desired contradiction.

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For positive numbers the function $f(x)=x^2$ is an increasing function. This follows from computing the derivative.

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    $\begingroup$ ... and follows more elementarily from the ordered field axioms: If $0<a<b$, then $a^2<ab<b^2$. $\endgroup$ – hmakholm left over Monica Feb 25 '17 at 0:45

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