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I have some questions about some things I want to clarify in regard to basic questions that ask to show that roots are irrational, for example $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{6}$. To me, I think there are a few little lemmas/thereoms I am using without explicility understanding, particularly in my assumptions about division. I will write some examples to clarify what I am wondering. I am looking for any advice, critiques, tips etc.

I am wanting to show that the following are irrational;

$\mathbf{\sqrt{3}}:$

The hint given for this one that I want to use is , to note that all integers can be written in one of the following ways, $3n$, $3n+1$ or $3n+2$.

Suppose that $\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ where $p,q \in \mathbb{Z}$ and $q \neq 0.$ We also can note that p and q are relatively prime, that is, there exists no common denominator. ( because if there was we could just divide by it and continue).

so we then have, $$\frac{p^2}{q^2}=3$$

giving $$p^2=3q^{2}$$

Now, how can I use that property of all numbers at this point? Or do I even need it? I am just not sure what to do that is valid, does this make any sense? from the property of all numbers we have that q is either of the form $3n$, $3n+1$ or $3n+2$ and hence $q^2$ must be of the form of one of $9n^2$, $9n^2+6n+2$ or $9n^2+12n+4$ but I'm not sure if I can make anything with this.

$\mathbf{\sqrt5}:$

With similar initial assumptions, we could write $p^{2}=5q^{2}$, which implies $p^2$ is divisible by $5$, and so does this imply p is divisible by 5 as well?

I think this is what I am looking for, something to do with if the square is divisible by some prime, what can we say about it?

Im sorry if what I am asking is confusing, I have more but I will leave it as is for now and update it soon. I hope it makes some sense , thanks!

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I think the hint is asking you to consider the equation:

$p^2 = 3q^2$ (mod $3$).

Since the right hand side is $0$ (mod $3$), so is the left, so this means $3|p^2 \implies 3|p$ (since $3$ is prime).

So say $p = 3t$. Then we have:

$9t^2 = 3q^2 \implies 3t^2 = q^2$. Taking this mod $3$ as well, we have:

$0 = q^2$ (mod $3$), so that $3|q^2 \implies 3|q$. Since we assumed $\gcd(p,q) = 1$, this is a contradiction.

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  • $\begingroup$ thanks. Can you possible elaborate on the terms, I understand that a=b mod n means n|a-b, so is what you wrote $3 | p^{2}-3q^{2}$? $\endgroup$ – Quality Jul 1 '15 at 4:28
  • $\begingroup$ If $3|(p^2 - 3q^2)$, by definition that means $p^2 - 3q^2 = 3k$ for some integer $k$, in which case $p^2 = 3(k + q^2)$, which is clearly divisible by $3$, yes? Even more directly: we see that $3q^2 - 0 = 3q^2$ so $3q^2 = 0$ (mod $3$). But $p^2$ EQUALS $3q^2$, so $p^2 - 0 = p^2$ is divisible by $3$ (it is $3$ times $q^2$). $\endgroup$ – David Wheeler Jul 1 '15 at 11:16
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All you have to do is to make a contradiction. So you can just begin with a discussion on whether $p$ and $q$ are multiples of $3$ (or $5$).

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It's a theorem that if a prime number $p$ divides $ab$, (integers), then $p$ divides $a$ or $p$ divides $b$. Now if $p$ divides $a^{2}$, then $p$ divides $a$ or $p$ divides $a$ so $p$ divides $a$. Edited:Now you can apply this as $5$ is prime, further, if a squarefree integer divides $a^{2}$, then it also divides $a$.

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  • $\begingroup$ So square free or prime works? $\endgroup$ – Quality Jun 30 '15 at 23:38
  • $\begingroup$ Yes!${}{}{}{}{}$ $\endgroup$ – mich95 Jun 30 '15 at 23:45
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You can also argue this... For any integer, $n$, and for any prime number, $p$, $n^2$ will contain an even number of $p's$ as factors. For example, $12^2 = (2^2\cdot3)^2 = 2^4 \cdot 3^2$. So there are four $2's$, two $3's$, zero $5's$, zero $7's$, and so on. So if $p^2 = 3q^2$, there there will be an even number of $3's$ on the left side and an odd number of $3's$ on the right side. This can't be true because then $p^2$ would factor into primes two distinct ways.

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