Proving irrationality

[Oaky]

...of \(\displaystyle \sqrt{3045}\)

Part A asked to prove that if \(\displaystyle m \in N\) and \(\displaystyle m^2\) is divisible by 3 then \(\displaystyle m\) is divisible by 3. (DONE)
Part B asked to hence prove that \(\displaystyle \sqrt{3}\) is irrational. (DONE)

Now apparently we can use part B to prove part C, as 3045 is a multiple of 3. But I have no idea how to do this.

Apparently the method mirrors the proof that \(\displaystyle \sqrt{15}\) is irrational.

I tried setting up some form of proof by contradiction by splitting \(\displaystyle \sqrt{3045}\) into \(\displaystyle \sqrt{3} \sqrt{1045}\) and assuming it equals \(\displaystyle p/q\) (no common factors).
So we get \(\displaystyle \sqrt{3} = p/\sqrt{1045}q\)... and I'm stuck. Where do I go from here?

It's also confusing because the proof obviously can't work for all multiples of 3, such as 9 or 36 etc.

Thanks so much

 

[pappus]

...of \(\displaystyle \sqrt{3045}\)

Part A asked to prove that if \(\displaystyle m \in N\) and \(\displaystyle m^2\) is divisible by 3 then \(\displaystyle m\) is divisible by 3. (DONE)
Part B asked to hence prove that \(\displaystyle \sqrt{3}\) is irrational. (DONE)

Now apparently we can use part B to prove part C, as 3045 is a multiple of 3. But I have no idea how to do this.

Apparently the method mirrors the proof that \(\displaystyle \sqrt{15}\) is irrational.

I tried setting up some form of proof by contradiction by splitting \(\displaystyle \sqrt{3045}\) into \(\displaystyle \sqrt{3} \sqrt{1045}\) and assuming it equals \(\displaystyle p/q\) (no common factors).
So we get \(\displaystyle \sqrt{3} = p/\sqrt{1045}q\)... and I'm stuck. Where do I go from here?

It's also confusing because the proof obviously can't work for all multiples of 3, such as 9 or 36 etc.

Thanks so much

If \(\displaystyle \sqrt{3} \cdot \sqrt{1045}\) is a rational number you can express it as a quotient of natural numbers with no common factors:

\(\displaystyle \displaystyle{\frac pq = \sqrt{3} \cdot \sqrt{1045}~\implies~\frac{p^2}{q^2} = 3 \cdot 1045}\) That means:

\(\displaystyle \displaystyle{p^2 = 3 \cdot 1045 \cdot q^2}\)

Obviously \(\displaystyle p^2\) is divisible by 3. So you can express p = 3k (that refers to part A of your question!):

\(\displaystyle (3k)^2 = 3 \cdot 1045 \cdot q^2~\implies~ 3k^2 = 1045 \cdot q^2\)

Obviously \(\displaystyle q^2\) is divisible by 3.

... and so on. That means you'll get a contradiction that there don't exist common factors. Since this premise is false the whole statement is false too.

 

[pappus]

This very first step was beyond me \(\displaystyle 3 * 1045 = 3(1000 + 40 + 5) = 3000 + 120 + 15 = 3135\ magically \equiv 3045.\)

I used Oaky's factorization without prior control. That will never (!) happen again

...
Lovely proof, Thanks! but one question. I do not get the "and so on."

Actually you can show by this procedure that p and q are not only divisible by 3 but by \(\displaystyle 3^n\) where n is the number of repetitions.

You assumed initially that p and q had no common factors (other than 1 of course). Once you show from this that both are divisible by 3, do you not have a sufficient contradiction? What more is left to do? Sorry to be dense.

Well, that's what I intended to say. Obviously my German grammar led to this confusion of understanding.

... and btw I like your being dense.

 

[Oaky]

\(\displaystyle \dfrac{3045}{3} = 1045\) was a herpaderp on my behalf, sorry!

The \(\displaystyle 3 \cdot k^2 = 1045 \cdot q^2 \implies q^2 \parallel 3\) was slightly confusing to begin with, but now I've wrapped my head around it.

Thanks so much everyone! Now off to prove the Binomial Therem, fun...