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



Note by Pi Han Goh
2 months 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.

- 3 weeks, 4 days 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 - 3 weeks, 4 days ago

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

- 1 month, 2 weeks ago

Edit: Ah yes, it is less than 4.

Staff - 3 weeks, 5 days ago

Comment deleted 3 weeks ago

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

- 1 month, 2 weeks ago

Is my proof necessarily wrong?

- 1 month, 2 weeks ago

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 - 3 weeks, 5 days ago

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

- 3 weeks, 4 days ago

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 - 3 weeks, 4 days ago

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

- 3 weeks, 3 days 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 $$\sqrt[3]{536}-\sqrt{17} \leq 4$$

- 3 days, 18 hours ago

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

Staff - 3 days, 17 hours ago