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I was doing some exercises from the book called Basica Mathematics - Serge Lang and here's the two problems: 1) Prove that there is no positive rational number a such that a^2 = 3. You may assume that a positive integer can be written in one of the forms 3k, 3k + 1, 3k + 2 for some integer k.

2) Prove that if the square of a positive integer is divisible by 3, then so is the integer. Then use a similar proof as for the square root of 2.

Basically I was trying to do this problems but I normally guide with the answers to see on how to approach this problems. If anybody has any idea on how to do this I would appreciate it

bl4ckch4ins
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1 Answers1

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Assume there exists such a number, $\frac{a}{b}$, with $a$ and $b$ integers. We can assume $a$ and $b$ have no common prime factors - if they do, divide them out to get a simplified fraction. For example, if we have $\frac{4}{2}$, the denominator and numerator share a common factor $2$; divide both by $2$ to get the fraction $\frac{2}{1}$.

Now the square, $\frac{a^2}{b^2}$, is equal to $\frac{3}{1}$. This is equivalent to $a^2 \cdot 1 = b^2 \cdot 3$. $3$ therefore is a divisor of $a^2$, and as $3$ is prime, this implies that $3$ divides $a$ (do you see why?). Ergo we can write $a = 3 \cdot q$, with $q$ another integer. Try to continue from there, substituting $a$ with $3 \cdot q$ in the original fraction, and conclude that $3$ also divides $b$. Why is this a contradiction?

Alexander Geldhof
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