Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$?

Question

Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$?

My thoughts: $\sqrt{2}$ is $\approx1,4$ and $\sqrt{3}$ is $\approx{1,7}$ so their sum must be of the interval $(3;4)$. Any more strict approaches? Thank you in advance!

Answer 1

First we have $\sqrt{2} + \sqrt{3} < \sqrt{4} + \sqrt{4} = 4$. To show that $\sqrt{2} + \sqrt{3} > 3$, we square both sides, and see that it suffices to show that $5 + 2\sqrt{6} > 9$. But this follows from $\sqrt{6} > 2$.

Answer 2

You are entirely right. To be more precise, you could say that $$ 1.4<\sqrt2<1.5\\ 1.7<\sqrt3<1.8 $$ and therefore $$ 3.1<\sqrt2+\sqrt3<3.3 $$ Alternately, to keep all approximations integral (and this is more likely the intended solution), we could also study $(\sqrt2+\sqrt3)^2=2+2\sqrt6+3$. Noting that $2<\sqrt6<3$ gives $$ 9<2+2\sqrt6+3<11<16 $$ which again means $3<\sqrt2+\sqrt3<4$.

Answer 3

I think they expect you to do something like this:

Since $2,3<4$ we know $\sqrt2,\sqrt3<2$, so $\sqrt2+\sqrt3<4$.

On the other hand, $(\sqrt2+\sqrt3)^2=2+3+2\sqrt6>9$ since $\sqrt6>\sqrt4=2$. This shows that $\sqrt2+\sqrt3>3$ so we have $$ 3<\sqrt2+\sqrt3<4$$

Answer 4

This first step (bracketing between plausible consecutive integers) was necessary.

Now we have to prove in a rigorous way that

$$\sqrt{2}+\sqrt{3}>3 \tag{1}$$ and

$$\sqrt{2}+\sqrt{3}<4 \tag{2}$$

Let us establish (2).

As function $x\to x^2$ is strictly increasing on $(0,\infty)$, therefore bijective, (2) is equivalent to:

$$(\sqrt{2}+\sqrt{3})^2<4^2 \ \iff \ 5+2 \sqrt{6}<16 \ \iff \ \sqrt{6}<\frac{11}{2} $$

itself equivalent, for the same reason, to:

$$6<\frac{121}{4} \ \iff 24 \ < \ 121$$

which is true.

Proceed in the same way for inequation (1).