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

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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.

Patrick Corn - 3 weeks, 4 days 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 - 3 weeks, 4 days ago

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

Krishnaraj Sambath - 1 month, 2 weeks ago

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Edit: Ah yes, it is less than 4.

Calvin Lin Staff - 3 weeks, 5 days ago

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Comment deleted 3 weeks ago

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

Pi Han Goh - 1 month, 2 weeks ago

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Is my proof necessarily wrong?

Deva Craig - 1 month, 2 weeks ago

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@Deva Craig 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.

Calvin Lin Staff - 3 weeks, 5 days ago

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

Deva Craig - 3 weeks, 4 days ago

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@Deva Craig 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.

Calvin Lin Staff - 3 weeks, 4 days ago

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

Deva Craig - 3 weeks, 3 days 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 - 3 days, 18 hours ago

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

Calvin Lin Staff - 3 days, 17 hours ago

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