# Prove $\sqrt{2}$ is irrational using prime number properties

I found the following argument in a textbook that uses a number theory approach. However I do not understand the last two sentences that seems to use some number theory properties, can someone please help explain them?

Let $$a^2 = 2b^2, \quad (a,b)=1, \quad a,b\in \mathbb{N}$$

Rearranging, $$b^2 = (a+b)(a-b)$$

Let $$p$$ be a prime factor of $$b$$, then $$p \mid (a+b)\quad or\quad p\mid (a-b)$$

If $$p$$ divides any of the above, then $$p$$ divides both of them, hence $$p$$ must divide $$a$$. Then $$p$$ would be a common divisor of $$a$$ and $$b$$, a contradiction.

• If $p | b$ and $p | (a-b)$, say, then $p |( b + (a-b))$ so $p | a$. Basically, you need the property that if $p$ divides two integers, then it also divides their sum and differences. – Malkoun Jan 18 '20 at 5:56
• If p is a factor of b, since $b^2 = (a+b)(a-b)$, that means that the factor p either came from $(a+b)$ or $(a-b)$, correct? Since p is a factor of b already, and hence of either $(a+b), (a-b)$ then it must also be a factor of a – Dhanvi Sreenivasan Jan 18 '20 at 6:01

(a+b) - (a-b) =2b ,so as their difference is divisible by p(it is obviuosly divisble by b and hence p), if any one (a-b) or (a+b) then other should also be divisible , hence their sum is divisble by p , sum is 2a.

• Huh??? I don't see how this solves the problem. – David G. Stork Jan 18 '20 at 6:09
• we have assumed that a and b are co primes, so if a prime factor of b ,we have proved it is a prime factor of a also and hence contradiction – aryan bansal Jan 18 '20 at 6:12
• There is a language/grammar problem here: "we have assumed that a and b are co primes, so if a prime factor of b ,we have proved it is a prime factor of a also and hence contradiction." Huh? "so if a prime factor of b"....is what??? or so if what? is a prime factor of b??? – David G. Stork Jan 18 '20 at 6:24
• i meant p is a prime factor of b – aryan bansal Jan 18 '20 at 9:51

The conclusion that $$p\mid(a+b)$$ or $$p\mid(a-b)$$ comes from the property of primes that if $$p\mid(rs)$$ then $$p\mid r$$ or $$p\mid s$$. This, in fact, is the defining property of prime elements in a ring.

For the next sentence: if $$p$$ divides $$a+b$$ then it divides $$(a+b)-2b=a-b$$; if $$p$$ divides $$a-b$$ then it divides $$(a-b)+2b=a+b$$. The author didn't need this step to conclude that $$p$$ divides $$a$$, however; they could have simply proceeded as in Malkoun's comment: if $$p$$ divides $$a+b$$ then $$p$$ divides $$(a+b)-b=a$$; if $$p$$ divides $$a-b$$ then $$p$$ divides $$(a-b)+b=a$$.

The final sentence of the proof just uses the assumption that $$p\mid b$$ and the conclusion that $$p\mid a$$ to infer that $$p$$ is a common divisor of $$a$$ and $$b$$, which violates the earlier assumption $$(a,b)=1$$.

Assume $$\sqrt{2} = \frac{a}{b}$$ with $$\{ a, b \} \in \mathbb{N}$$ and reduced form. Then $$2 b^2 = a^2 .$$

The number of prime factors on the left is odd; the number of prime factors on the right is even. From the fundamental theorem of arithmetic (unique prime factorization), this cannot occur.

QED

• My goodness! I would love to hear from the downvoter why this (which I learned in math at MIT) is not correct. – David G. Stork Dec 4 '20 at 4:10