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Which of the following best explains why $\sqrt{3}$ is irrational?
ABecause 3 is not a natural number
BBecause assuming $\sqrt{3}=\frac{a}{b}$ in lowest terms forces both $a$ and $b$ to be divisible by 3
CBecause $\sqrt{3}$ has a terminating decimal expansion
DBecause every square root is irrational
Answer & Solution
Correct answer: B. Because assuming $\sqrt{3}=\frac{a}{b}$ in lowest terms forces both $a$ and $b$ to be divisible by 3
Assume $\sqrt{3}=\frac{a}{b}$ with $a$ and $b$ coprime. Then $3b^2=a^2$, so 3 divides $a^2$, hence 3 divides $a$. Substituting $a=3c$ gives $b^2=3c^2$, so 3 divides $b$ too. This contradicts $a$ and $b$ being coprime, so $\sqrt{3}$ is irrational.
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