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How do I guess which number is greater by using logic? Are there any tricks for guessing?

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  • $\begingroup$ note that $-2 = \sqrt{4} > \sqrt{9} = -3$ $\endgroup$ – reuns Sep 2 '16 at 4:05
  • $\begingroup$ @user1952009 $\sqrt 4$ is not equal to $-2$ by any means! $\endgroup$ – Babai Sep 2 '16 at 4:37
  • $\begingroup$ @user1952009: note that $\sqrt x$ is a nonnegative number by definition (assuming nonnegative $x$). $\endgroup$ – MPW Sep 2 '16 at 4:49
  • $\begingroup$ @MPW it was a joke... $\endgroup$ – reuns Sep 2 '16 at 4:49
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    $\begingroup$ @MPW I was trying to be fun $\endgroup$ – reuns Sep 2 '16 at 4:52
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Hint: Observe that $\sqrt{3}-\sqrt{2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}$.

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  • $\begingroup$ This reasoning is circular because it either assumes that $\sqrt{3}, \sqrt{2} > \frac{1}{2}$ or that $\sqrt{2}, \sqrt{3} > \sqrt{1}$. $\endgroup$ – Jared Sep 2 '16 at 4:33
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    $\begingroup$ @Jared: Huh? The RHS is a positive number, so the minuend on the LHS must be greater than the subtrahend. This is an excellent proof. $\endgroup$ – MPW Sep 2 '16 at 4:43
  • $\begingroup$ @MPW You are correct, I see my mistake now. $\endgroup$ – Jared Sep 2 '16 at 5:05
  • $\begingroup$ I cannot undo my erroneous downvote unless an edit is made...if so, I will, but otherwise take this comment as adding two up-votes to this answer. $\endgroup$ – Jared Sep 2 '16 at 5:08
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    $\begingroup$ @Jared: No need for that. I am glad that you considered leaving a comment with your downvote. $\endgroup$ – user 170039 Sep 2 '16 at 5:20
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It's extremely easy. Note that the square root function is strictly increasing across the non-negative real numbers. Since $3>2$ it follows that $\sqrt3>\sqrt2$.

In general:

  • Any function that is strictly increasing, like $\ln x$, will always preserve the direction of an inequality when it is applied to both sides.
  • Any function that is strictly decreasing, like $e^{-x}$, will always flip the inequality's direction.
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Note that the function $f(x)=\sqrt{x}$ is defined on the set $(0,\infty)$, and that on that range, the derivative $$f'(x)=\frac{1}{2\sqrt{x}}$$ is always $>0$. Therefore $f$ is strictly increasing, so in fact $\sqrt{a}< \sqrt{b}\iff a<b$ for any two $a,b\in(0,\infty)$.

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If $0 <b $ and $x <y $ then $bx <by $. That's an axiom.

So if $a < b $ then $a^2 = a*a <a*b <b*b = b^2$

We can conclude that $\sqrt {a} < \sqrt {b} $ because if it weren't then $a = \sqrt {a}^2 < \sqrt {b}^2=b $ wouldn't be true. (If $\sqrt {a} \ge \sqrt {b} $ then $\sqrt {a}^2 \ge \sqrt {b}^2$.)

So $\sqrt {2} < \sqrt {3}$ simply because $2 <3$.

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