Use your imagination!

Do you have any problems that can be solved very easily with an alternative solution using very high-level theorems? I have added one below.

Note by Shenal Kotuwewatta
3 years ago

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  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

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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.

Pi Han Goh - 3 years ago

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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?

Pi Han Goh - 3 years ago

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@Pi Han Goh Nope. What a coincidence. xD

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 $1$.

Shenal Kotuwewatta - 3 years ago

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