5
$\begingroup$

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.

  • 7
    If $a>b>0$ then $a^2>b^2$. It is not true if $b<0$. – 2017-02-25
  • 0
    Consider a right triangle with legs $1$ and $2$. Then $\sqrt{5}=c\leq a+b=3$. – 2017-02-25

5 Answers 5

9

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.

  • 1
    "Since $3<\sqrt{5}$"? Sorry, that can't be right, unless you change it into "Since we supposed $3<\sqrt{5}$..... – 2017-02-25
  • 2
    @imranfat the answerer is assuming that $\sqrt{5}>3$. – 2017-02-25
  • 1
    @Xam, yes, it just looks odd... – 2017-02-25
3

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)

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

$\sqrt 5 <3$.

Q.E.D.

  • 0
    Thank you for this, unnecessary contradiction proofs are so ugly. – 2017-02-25
  • 0
    How would I prove the square-root function is increasing? I don't like making assumptions like that. – 2017-02-25
  • 1
    @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) – 2017-02-26
2

Hint: $\;\;3-\sqrt{5} \,=\, \cfrac{4}{3+\sqrt{5}}$

1

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.

0

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

  • 5
    ... and follows more elementarily from the ordered field axioms: If $0 – 2017-02-25