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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
This can be proved by contradiction. Suppose 2+3 is rational. Then 2+31=3−2 is rational as well. Now since the sum of rational numbers is always rational, we would then have that
21((2+3)+(3−2))=3
is rational, which is not the case. So our original assumption must be false, i.e., 2+3 is irrational.
Note that we can prove 3 is irrational by contradiction. Suppose that 3=ba for positive coprime integers a,b. Squaring both sides gives us that 3b2=a2, which by the Fundamental Theorem of Arithmetic implies that 3∣a. But if a=3m then 3b2=9m2⟹b2=3m2, implying that 3∣b. But we had assumed that a,b were coprime, and thus our original assumption that 3 was rational is false.
Note that 2,3 are both algebraic integers, so 2+3 is also an algebraic integer. However, if 2+3 is rational then that means it is a rational integer, which is clearly false as 3<2+3<4.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
This can be proved by contradiction. Suppose 2+3 is rational. Then 2+31=3−2 is rational as well. Now since the sum of rational numbers is always rational, we would then have that
21((2+3)+(3−2))=3
is rational, which is not the case. So our original assumption must be false, i.e., 2+3 is irrational.
Note that we can prove 3 is irrational by contradiction. Suppose that 3=ba for positive coprime integers a,b. Squaring both sides gives us that 3b2=a2, which by the Fundamental Theorem of Arithmetic implies that 3∣a. But if a=3m then 3b2=9m2⟹b2=3m2, implying that 3∣b. But we had assumed that a,b were coprime, and thus our original assumption that 3 was rational is false.
Note that 2,3 are both algebraic integers, so 2+3 is also an algebraic integer. However, if 2+3 is rational then that means it is a rational integer, which is clearly false as 3<2+3<4.
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Extension: if a1+a2+⋯+an is rational, prove that it is integer.
In particular, prove that a1,a2,…,an are perfect squares.