# Prove that $\sqrt{536} - \sqrt{17} < 4$ without using a calculator Note by Pi Han Goh
3 years, 11 months ago

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Well... \begin{aligned} 71824 &< 71825 \quad \quad \text{(take square roots)} \\ 268 &< 65 \sqrt{17} \\ 536 &< 268 + 65 \sqrt{17} \quad \quad \text{(take cube roots)} \\ \sqrt{536} &< 4+\sqrt{17} \\ \sqrt{536} - \sqrt{17} &< 4. \end{aligned} It's not enlightening but it works (and getting this without using a calculator is a little tedious, but not crazy).

The inequality at the beginning is pretty tight! The actual value is only very slightly less than 4. But anyway, note the direction of the inequality, contrary to the statement of the problem.

- 3 years, 9 months ago

Nice. I've edited the title accordingly.

Where the number of terms is small (less than 4-6), we can repeatedly square and cube things to get rid of the surds.

FYI Reversing the direction would make more sense as how someone would approach it. IE:

\begin{aligned} & \sqrt{536} - \sqrt{17} < 4 \\ \Leftrightarrow & \sqrt{536} < 4 + \sqrt{17} \\ \Leftrightarrow & 536 <268 + 65 \sqrt{17} \\ \Leftrightarrow & 268 < 65 \sqrt{17} \\ \Leftrightarrow & 71824 < 71825 \\ \end{aligned}

Staff - 3 years, 9 months ago

You have typed wrong! it should be 17 but not 18 on the fourth line of proof

- 1 year, 11 months ago

Edited, thanks.

Staff - 1 year, 11 months ago

Actually, using calculator, cuberoot(536) - sqrt(17) < 4..

- 3 years, 10 months ago

Edit: Ah yes, it is less than 4.

Staff - 3 years, 10 months ago

$4^2=16<17<25=5^2$, meaning that $4<\sqrt{17}<5$; we label this inequality as $\boxed{1}$. $8^3=216<536<729=9^3$, meaning that $8<\sqrt{536}<9$; we label this inequality as $\boxed{2}$. $\boxed{2}$-$\boxed{1}$ shows that $3 \leq \sqrt{536}-\sqrt{17} \leq 4$, meaning that $\sqrt{536}-\sqrt{17} \leq 4$

- 3 years, 9 months ago

Unfortunately, that's not how inequalities work. Do you see your mistake?

Staff - 3 years, 9 months ago