What is the most rigorous proof of the irrationality of the square root of 3?

Question

I am currently trying to self-study Stephen Abbott's Understanding Analysis. The first exercise asks to prove the irrationality of √3, and I understand the general idea of the contradiction by finding that the relatively prime integers p and q have a common factor. However, I am stuck on the idea that if p^2 is divisible by 3, then p is divisible by 3. Abbott's solution assumes this, but I have also seen proofs that analyze the situations where a and b are even or odd (such as NASA's). Even or odd really is just saying multiple of 2, which confuses me as to why the even/odd method (which is much less concise) would be used.

Sorry for the block of rambling text, I just want to start writing proofs the right way. I guess my real questions are:

If p^2 is divisible by a prime number, is p also divisible by that prime number? Can this just be assumed, or is there a theorem I have to mention in the proof? Why do some proofs analyze the even/odd situations of a and b? Are they more rigorous, and if they are not, why are they used, considering their added length and complexity? Finally, am I simply over thinking the idea of being rigorous and missing the big picture?

Answer 1

If a product $ab$ is divisible by a prime $p$, then at least one of the factors is divisible by $p$. If you do not already know this (and Abbott has not proved it prior to this point of the book), then you may need to go back to a more elementary book for its proof.

added
Euclid, Book VII, Propsition 30

Answer 2

Suppose that $\sqrt{3}$ is rational and $\sqrt{3}=a/b$ then squaring both sides gives you $3b^{2}=a^{2}$. We can suppose that $a/b$ is reduced (so $a$ and $b$ have no common factors). This means, in particular, that it is not the case that both $a$ and $b$ are even. Note that $n^{2}\equiv 0,1\mod 4$, $n^2\equiv 0\mod 4 $ if $n$ is even and $1$ otherwise. Then, look at $3b^{2}\equiv a^{2}\mod 4$. It is inconsistent. Contradiction.

As for the second (tangent) question: if $p|n^{2}$ where $p$ is a prime, then it is true that $p|n$. Think about the prime factorization of $n^{2}$... $n^{2}=(n)^{2}=(p_1^{k_1}...p_{m}^{k_m})^{2}=p_{1}^{2k_1}...p_{m}^{2k_m}$. So $n$ and $n^2$ have exactly the same prime divisors and knowing how $n$ (or $n^2$) factors tells you how $n^{2}$ (or $n$) factors.

Answer 3

Prompted by the title of the question, FYI here is a sketch of another very nice proof.

If $\sqrt3$ is equal to the fraction $$\frac pq$$ which is a quotient of positive integers, then it is also equal to $$\frac{3q-p}{p-q}\ ,$$ which is a quotient of positive integers with smaller denominator.

Answer 4

Lemma: $a^2$ is divisible by $3$ if and only if $a$ is.

Proof: By examining all possible cases:

• If $a$ is divisible by $3$, $a^2$ is, trivially.
• If $a$ is not divisible by $3$, $a^2$ is not, either.

Indeed, if $a$ is not divisible by $3$, then $a=3k\pm1$ for some $k$. Then $$a^2=9k^2\pm6k+1=3(3k^2\pm2k)+1$$ has remainder $1$ upon division by $3$.

Answer 5

personally I prefer to prove these results by contrapositive. If $p,q$ are coprime positive integers with $q>1$ then $$ p^{k}/q^{k} $$ is not an integer for any $k>0.$ This immediately implies the irrationality of all roots of $3.$ In fact, it proves that any root of an integer that is not an integer is irrational.

The key is, as everyone has already said, is that $p,q$ coprime implies that $p^{k}$ and $q^k$ are also coprime, and so $p^k/q^k$ is not an integer.

This is implied by that if a prime $m$ divides $ab$ then it divides $a$ or $b.$ So if it divides $p^k$ it divides $p$ and if it divides $q^k$ it divides $q.$ So no prime will divide both of them unless it divides $p$ and $q$ which we ruled out.

How do we prove the result that $m$ must divide $a$ or $b$? Either it divides $a$ and we are done or $m$ and $a$ are co-prime since $m$ is prime. We then have that there exist $\alpha $ and $\beta$ such that $$ \alpha m + \beta a =1. $$ (this follows from the Euclidean algorithm.) So $$ b\alpha m + \beta ab =b $$ Since $m$ divides $ab$ it divides the LHS, so it divides the RHS too.

(See my book "proof patterns" for more discussion.)

Answer 6

Let us try a new one. Assume $a = \sqrt{3}b$ for integers a and b, then $a^2 = 3b^2$. Now use uniqueness of prime factorization and count the occurences the prime 3 has in the numbers. if $a = 3^{d}c$ and $b = 3^{g}f$, then we have that $a^2 = 3^{2d}c^2$ and that $3b^2 = 3^{2g+1}f^2$.

We now see that left hand side has even number of 3-factors but right hand side has odd - contradiction.

Answer 7

Here is a very simple general proof that the square root of any integer that is not a perfect square is irrational. For convenience, write $a|b$ (read "a divides b") for "b is divisible by a". The proof depends on Euclid's Lemma, which states that if $gcd(a, b) = 1$ and $a | bc$, then $a | c$.

For positive integers $a$, $b$, and $n$, assume $(\frac a b)^2 = n$, where $gcd(a, b) = 1$ (i.e., the fraction is in lowest terms). Then $a^2 = b^2n$, so $a | (b^2n)$.

By Euclid's Lemma, $a | bn$, and applying the Lemma again, $a | n$. Since $n$ is divisible by $a$, we may divide both sides of $a^2 = b^2 n$ by $a$ to get $a = b^2(\frac n a)$, where $(\frac n a)$ is an integer.

But this means that $b | a$, so $gcd(a, b) = b$. Since we know $gcd(a, b) = 1$, this means that $b = 1$, which implies, in turn, that $n = a^2$. That is, $n$ is the square of an integer, i.e., a perfect square.

We have just proven that any integer that is the square of a rational number is a perfect square. This is logically equivalent to the assertion that any integer that is not a perfect square is not the square of any rational number, i.e., the square root of any integer that is not a perfect square is irrational.