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.

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.

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

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## Comments

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

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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:

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

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

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Edit: Ah yes, it is less than 4.

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Comment deleted 6 months 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?

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Is my proof necessarily wrong?

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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.

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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.

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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 \(\sqrt[3]{536}-\sqrt{17} \leq 4\)

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

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