# Prove that $$\sqrt[3]{536} - \sqrt{17} < 4$$ without using a calculator



Note by Pi Han Goh
1 year, 1 month ago

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Well... \begin{align} 71824 &< 71825 \quad \quad \text{(take square roots)} \\ 268 &< 65 \sqrt{17} \\ 536 &< 268 + 65 \sqrt{17} \quad \quad \text{(take cube roots)} \\ \sqrt[3]{536} &< 4+\sqrt{17} \\ \sqrt[3]{536} - \sqrt{17} &< 4. \end{align} 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.

- 1 year 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{align} & \sqrt[3]{536} - \sqrt{17} < 4 \\ \Leftrightarrow & \sqrt[3]{536} < 4 + \sqrt{17} \\ \Leftrightarrow & 536 <268 + 65 \sqrt{17} \\ \Leftrightarrow & 268 < 65 \sqrt{18} \\ \Leftrightarrow & 71824 < 71825 \\ \end{align}

Staff - 1 year ago

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

- 1 year, 1 month ago

Edit: Ah yes, it is less than 4.

Staff - 1 year ago

Comment deleted Sep 28, 2017

So, you have established that 4<sqrt17 <5 and 8< cbrt536<9, so why is the inequality in question fulfilled?

- 1 year, 1 month ago

Is my proof necessarily wrong?

- 1 year, 1 month ago

Yes, your proof is wrong.

You are saying that if $$8 < c < 9$$ and $$8 < a + b < 9$$, then we can conclude that $$c > a + b$$. There is no reason why this must be true.

Staff - 1 year ago

I don't quite understand what you mean. Is it alright if I delete my proof and try to improve it?

- 1 year ago

Yup, go ahead and restart.

All that you have shown is that both $$\sqrt[3]{536}$$ (in step 4) and $$4 + \sqrt{17}$$ (in step 3) are between 8 and 9. You have not shown how these 2 numbers compare to each other.

Staff - 1 year ago

I deleted the proof and copy and pasted it onto a Word document.

- 1 year 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 $$8<\sqrt[3]{536}<9$$; we label this inequality as $$\boxed{2}$$. $$\boxed{2}$$-$$\boxed{1}$$ shows that $$3 \leq \sqrt[3]{536}-\sqrt{17} \leq 4$$, meaning that $$\sqrt[3]{536}-\sqrt{17} \leq 4$$

- 12 months ago

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

Staff - 12 months ago

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