Problem 1: Prove that \(\sqrt[3]{7} \) is an irrational number.

Conventional solution:

Prove by Rational Root Theorem. Suppose otherwise, then let \(x= \sqrt[3]{7} \) for rational \(x\), equivalently, we have \(x^3 - 7 = 0 \). By rational root theorem, we have \(\pm 1, \pm 7\) as possible rational solution, however, by trial and error, we can see that none of these values satisfy the equation, which is absurd. Thus, it must be irrational.

Overkill solution:

Prove by Fermat's Last Theorem. Suppose otherwise, then let \(\sqrt[3]{7} = \frac pq \) for coprime positive integers \(p,q\). Then \(7 = \frac{p^3}{q^3} \Rightarrow 7q^3 = p^3 \), or \(8q^3 = p^3 + q^3 \Rightarrow (2q)^3 = p^3 + q^3 \) which contradicts Fermat's Last Theorem.

* Problem 2: * Prove that \( \sqrt[n]{2} \) is an irrational number for \( n > 2 \).

* Overkill solution: *

Proof by Fermat's last theorem. Suppose otherwise, then let \( \sqrt[n]{2} = \frac{p}{q} \implies 2 = \frac{p^n}{q^n} \), or \( 2q^n = p^n \implies q^n + q^n = p^n \), which contradicts Fermat's last theorem.

I don't really use the wikis. I did see this somewhere on Brilliant though. Much before the wikis. Maybe it was one of Sreejato's questions. I'm not sure TBH.

Prove that the sequence \[1,11,111,...\] contains no perfect squares other than \(1\).

Solution 1:

Notice that every term after the first term is congruent to \(11 \equiv 3 \pmod 4\). But no square is congruent to \(3 \pmod 4\), so no term after the first term is a perfect square.

Solution 2:

Observe that

\[111...11=\frac{999...99}{9}=\frac{10^n-1}{9}\]

for some \(n \in Z. (n>1)\)

Assume that

\[\frac{10^n-1}{9}=k^2\].

Then

\[10^n-1=9k^2=(3k)^2\]

so

\[10^n=(3k)^2+1\].

But since \(n>1\), by Mihailescu's Theorem, we get a contradiction.

So there are no perfect squares in the sequence except $1$.

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## Comments

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TopNewestProblem 1: Prove that \(\sqrt[3]{7} \) is an irrational number.Conventional solution:Prove by Rational Root Theorem. Suppose otherwise, then let \(x= \sqrt[3]{7} \) for rational \(x\), equivalently, we have \(x^3 - 7 = 0 \). By rational root theorem, we have \(\pm 1, \pm 7\) as possible rational solution, however, by trial and error, we can see that none of these values satisfy the equation, which is absurd. Thus, it must be irrational.

Overkill solution:Prove by Fermat's Last Theorem. Suppose otherwise, then let \(\sqrt[3]{7} = \frac pq \) for coprime positive integers \(p,q\). Then \(7 = \frac{p^3}{q^3} \Rightarrow 7q^3 = p^3 \), or \(8q^3 = p^3 + q^3 \Rightarrow (2q)^3 = p^3 + q^3 \) which contradicts Fermat's Last Theorem.

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Kind of similar, so adding this as a comment.

* Problem 2: *Prove that \( \sqrt[n]{2} \) is an irrational number for \( n > 2 \).* Overkill solution: *Proof by Fermat's last theorem. Suppose otherwise, then let \( \sqrt[n]{2} = \frac{p}{q} \implies 2 = \frac{p^n}{q^n} \), or \( 2q^n = p^n \implies q^n + q^n = p^n \), which contradicts Fermat's last theorem.

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Hey, did you stumble upon this Wiki by accident?

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I don't really use the wikis. I did see this somewhere on Brilliant though. Much before the wikis. Maybe it was one of Sreejato's questions. I'm not sure TBH.

Sorry for copying your proof.

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Problem:Prove that the sequence \[1,11,111,...\] contains no perfect squares other than \(1\).

Solution 1:Notice that every term after the first term is congruent to \(11 \equiv 3 \pmod 4\). But no square is congruent to \(3 \pmod 4\), so no term after the first term is a perfect square.

Solution 2:Observe that

\[111...11=\frac{999...99}{9}=\frac{10^n-1}{9}\]

for some \(n \in Z. (n>1)\)

Assume that

\[\frac{10^n-1}{9}=k^2\].

Then

\[10^n-1=9k^2=(3k)^2\]

so

\[10^n=(3k)^2+1\].

But since \(n>1\), by Mihailescu's Theorem, we get a contradiction.

So there are no perfect squares in the sequence except $1$.

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