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

\(\)

Note by Pi Han Goh
1 year, 5 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.

Patrick Corn - 1 year, 3 months ago

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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:

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

Calvin Lin Staff - 1 year, 3 months ago

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

Krishnaraj Sambath - 1 year, 4 months ago

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Edit: Ah yes, it is less than 4.

Calvin Lin Staff - 1 year, 3 months ago

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\(4^2=16<17<25=5^2\), 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 \(\sqrt[3]{536}-\sqrt{17} \leq 4\)

Ken Kai - 1 year, 3 months ago

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

Calvin Lin Staff - 1 year, 3 months ago

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