MHB Proving the Irrationality of $\sqrt{3}$

[paulmdrdo1]

prove that $\sqrt{3}$ is irrational.

this is what I tried

$\sqrt{3}=\frac{p}{q}$ whee p and q are integers in lowest terms. common factor of +\-1 only.

squaring both sides

$\frac{p^2}{q^2}=3$

$p^2=3q^2$ assuming that $3q^2$ is even then $p^2$ is even hence p is also even.

$(3k)^2=3q^2$ where is k is an even integer.

then,

$q^2=3k^2$ q is also even

if p and q is even they have common factor of 3, thus contradicting the assumption that they have no common factor except +\-1. hence $\sqrt{3}$ is irrational.

is my proof correct? if it's not, can you show me the correct proof of this.

thanks!(Talking)

 

[Opalg]

prove that $\sqrt{3}$ is irrational.

this is what I tried

$\sqrt{3}=\frac{p}{q}$ whee p and q are integers in lowest terms. common factor of +\-1 only.

squaring both sides

$\frac{p^2}{q^2}=3$

$p^2=3q^2$ assuming that $3q^2$ is even then $p^2$ is even hence p is also even.

$(3k)^2=3q^2$ where is k is an even integer.

then,

$q^2=3k^2$ q is also even

if p and q is even they have common factor of 3, thus contradicting the assumption that they have no common factor except +\-1. hence $\sqrt{3}$ is irrational.

is my proof correct? if it's not, can you show me the correct proof of this.

thanks!(Talking)

You have the right idea, but it is buried in a layer of confusion surrounding the word "even". You are obviously trying to model your proof on the standard proof that $\sqrt2$ is irrational. The proof for $\sqrt2$ is based on showing that in that case $p$ and $q$ must both be even. But the reason that even numbers are important in that proof is that an even number is a multiple of $2$. If you go back to that proof and substitute "multiple of $2$" every time you see the word "even", then you will have a better idea of how to do the proof for $\sqrt3$. All you will have to do is to replace "multiple of $2$" by "multiple of $3$". The proof for $\sqrt3$ should not involve the word "even" at all. What really matters (as you have realized towards the end of your argument) is that $p$ and $q$ must both be multiples of $3$. If you focus your argument entirely on that, and leave out all references to numbers being even, then you will have a good proof of this result.

 

[Prove It]

It would also be a good idea to prove that the square of number is only a multiple of three if the number was a multiple of three.

This would be easiest to prove using the contrapositive, which is "if the number is not a multiple of three, then the square is not a multiple of three".

So if we have integers p and k such that $\displaystyle \begin{align*} p = 3k + 1 \end{align*}$ then

$\displaystyle \begin{align*} p^2 &= \left( 3k + 1 \right) ^2 \\ &= 3k^2 + 6k + 1 \\ &= 3 \left( k^2 + 2k \right) + 1 \end{align*}$

which is not a multiple of three. Also if p and k were integers such that $\displaystyle \begin{align*} p = 3k + 2 \end{align*}$, then

$\displaystyle \begin{align*} p^2 &= \left( 3k + 2 \right) ^2 \\ &= 9k^2 + 12k + 4 \\ &= 9k^2 + 12k + 3 + 1 \\ &= 3 \left( 3k^2 + 4k + 1 \right) + 1 \end{align*}$

which is also not a multiple of three.

Therefore if the square of a number was a multiple of three, the original number had to have been a multiple of three. Q.E.D.

 

[Deveno]

A lesser known form of proof runs as follows:

If $n$ is a natural number, and $\sqrt{n} \in \Bbb Q$, then $\sqrt{n} \in \Bbb Z$.

You may want to reflect a bit on why this is so, because it is NOT obvious. I suggest approaching this problem like so:

Suppose $n = \dfrac{a^2}{b^2}$, and that $p$ is a prime for which $p^k|b$ but $p^{k+1}\not\mid b$. Show that $p^k|a$, and thus conclude $b|a$.

So in looking for square root(s) of 3, we need only consider $\{-3,-2,-1,0,1,2,3\}$, which we can test individually.

In particular, any square-free positive integer must have an irrational square root (this is a powerful result, as it handles MANY cases "all at once").

 

[paulmdrdo1]

This is my second attemp

$\sqrt{3}=\frac{p}{q}$ whee p and q are integers in lowest terms. common factor of +\-1 only.

squaring both sides

$\frac{p^2}{q^2}=3$

$p^2=3q^2$ since $3q^2$ is a multiple of 3 p must also be a multiple of 3.

now I have,

$(3k)^2=3q^2$

then,

$q^2=3k^2$ $q^2$ is also a multiple of 3.

but if p and q is even they have common factor of 3, thus contradicting the assumption that they have no common factor except +\-1. hence $\sqrt{3}$ is irrational.

is this now correct?

thanks!(Talking)