Prove that ## \sqrt{p} ## is irrational for any prime ## p ##?

[Math100]

Homework Statement

Establish the following facts:
a) $\sqrt{p}$ is irrational for any prime $p$.

Relevant Equations

None.

Proof:

Suppose for the sake of contradiction that $\sqrt{p}$ is not irrational for any prime $p$,
that is, $\sqrt{p}$ is rational.
Then we have $\sqrt{p}=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ such that
$gcd(a, b)=1$ where $b\neq 0$.
Thus $p=\frac{a^2}{b^2}$,
which implies $pb^2=a^2$.
Note that $p\mid a^2$.
This means $p\mid a$, because $p$ is a prime number.
Now we have $a=pm$ for some $m\in\mathbb{Z}$.
Thus $a=pm$
$a^2=(pm)^2$
$pb^2=p^2 m^2$,
or $b^2=pm^2$.
Note that $p\mid b^2$.
This means $p\mid b$, because $p$ is a prime number.
Since $p\mid a\land p\mid b$,
it follows that $gcd(a, b)\neq 1$.
This is a contradiction because the integers $a$ and $b$ are relatively prime
with no common factors except $1$.
Therefore, $\sqrt{p}$ is irrational for any prime $p$.

 

[nuuskur]

Suppose for the sake of contradiction that $\sqrt{p}$ is not irrational for any prime $p$,

If the initial statement is "$\sqrt{p}$ is irrational for every prime $p$", then your supposition is not to the contrary. You are saying "suppose $\sqrt{p}$ is rational for every prime $p$".

What follows seems fine to me

Edit: Thought about it some more. I think, you mean to express the correct thing. Your sentence is to be read as "suppose $\sqrt{p}$ is $\neg$(irrational for every $p$)". To reduce ambiguity, you could express the same thought by negating the initial statement: suppose there exists a prime $p$ such that $\sqrt{p}$ is rational.
Alternatively, in the language of symbols, the initial statement is
$$\forall p\in\mathbb P,\quad \sqrt{p} \in \mathbb R\setminus \mathbb Q.$$
Hence its negation is
$$\exists p\in\mathbb P,\quad \sqrt{p} \in\mathbb Q.$$

 

[Math100]

If the initial statement is "$\sqrt{p}$ is irrational for every prime $p$", then your supposition is not to the contrary. You are saying "suppose $\sqrt{p}$ is rational for every prime $p$".

What follows seems fine to me

Edit: Thought about it some more. I think, you mean to express the correct thing. Your sentence is to be read as "suppose $\sqrt{p}$ is $\neg$(irrational for every $p$)". To reduce ambiguity, you could express the same thought by negating the initial statement: suppose there exists a prime $p$ such that $\sqrt{p}$ is rational.
Alternatively, in the language of symbols, the initial statement is
$$\forall p\in\mathbb P,\quad \sqrt{p} \in \mathbb R\setminus \mathbb Q.$$
Hence its negation is
$$\exists p\in\mathbb P,\quad \sqrt{p} \in\mathbb Q.$$

Besides "Suppose there exists a prime $p$ such that $\sqrt{p}$ is rational", do I also need to include the "Suppose for the sake of contradiction" to this first sentence or no?

 

[Mark44]

Besides "Suppose there exists a prime $p$ such that $\sqrt{p}$ is rational", do I also need to include the "Suppose for the sake of contradiction" to this first sentence or no?

No. If your supposition reaches a contradiction, then the supposition is false. That is, no such prime p exists for which its square root is rational.

 

[nuuskur]

do I also need to include the "Suppose for the sake of contradiction" to this first sentence or no?

It is not a necessity. We say things like "suppose for a contradiction that so and so.." to indicate the proof is a non-constructive one. With that said, if you are able to provide a direct proof to a statement, then that is preferable.

For beginners, I strongly recommend writing these things, however. It helps in the readability department. While I may not be a beginner, anymore, I am still clearly indicating in my proofs what sort of technique I use.

 

[mathwonk]

I think this is a very nice proof. One thing about it surprizes me however, namely that you use the technically correct mathematical definition of "prime". Namely that p is prime iff whenever p divides a product, then it divides one of the factors. In elementary mathematics, one often encounters the use of the word "prime", to mean rather "irreducible", namely an integer of absolute value greater than one, that is not a product of two other integers also of absolute value greater than one. So it is of interest to be able to prove here also that "irreducible" implies "prime". Otherwise you have a problem proving the existence of prime integers. I.e. it is obvious that 3 say, is irreducible, but not so obvious that 3 is prime in the strict mathematical sense. So I am asking if you can also prove that the square root of an irreducible integer is irrational. With what you have done, it would suffice to prove that irreducible implies prime. But perhaps you have already seen that in your course. The key is usually to use (and prove) the result that the largest (positive) integer dividing two others, equals the smallest (positive) integer that can be expressed as a linear combination of the two.