Novel (?) proof of the irrationality of $\sqrt3$ [duplicate]

I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if $x$ is divisible by $3$, then $x^2$ is divisible by $3$. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of $6$?

Solution 1:

Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.

So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

What is the contradiction?

Solution 2:

suppose $\sqrt{3}$ is rational, then $\sqrt{3}=\frac{a}{b} $ for some $(a,b) $ suppose we have $a/b$ in simplest form.
\begin{align} \sqrt{3}&=\frac{a}{b}\\ a^2&=3b^2 \end{align} if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.
if $b$ is odd then $a$ is also odd. Therefore:
\begin{align} a&=2n+1\\ b&=2m+1\\ (2n+1)^2&=3(2m+1)^2\\ 4n^2+4n+1&=12m^2+12m+3\\ 4n^2+4n&=12m^2+12m+2\\ 2n^2+2n&=6m^2+6m+1\\ 2(n^2+n)&=2(3m^2+3m)+1 \end{align} Since $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

Solution 3:

A supposed equation $m^2=3n^2$ is a direct contradiction to the Fundamental Theorem of Arithmetic, because when the left-hand side is expressed as the product of primes, there are evenly many $3$’s there, while there are oddly many on the right.

Solution 4:

The number $\sqrt{3}$ is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).

So the Assumptions states that :

(1) $\sqrt{3}=\frac{a}{b}$

Where a and b are 2 integers

Now since we want to disapprove our assumption in order to get our desired result, we must show that there are no such two integers.

Squaring both sides give :

$3=\frac{a^2}{b^2}$

$3b^2=a^2$

(Note : If $b$ is odd then $b^2$ is Odd, then $a^2$ is odd because $a^2=3b^2$ (3 times an odd number squared is odd) and Ofcourse a is odd too, because $\sqrt{odd number}$ is also odd.

With a and b odd, we can say that :

$a=2x+1$

$b=2y+1$

Where x and y must be integer values, otherwise obviously a and b wont be integer.

Substituting these equations to $3b^2=a^2$ gives :

$3(2y+1)^2=(2x+1)^2$

$3(4y^2 + 4y + 1) = 4x^2 + 4x + 1$

Then simplying and using algebra we get:

$6y^2 + 6y + 1 = 2x^2 + 2x$

You should understand that the LHS is an odd number. Why?

$6y^2+6y$ is even Always, so +1 to an even number gives an ODD number.

The RHS side is an even number. Why? (Similar Reason)

$2x^2+2x$ is even Always, and there is NO +1 like there was in the LHS to make it ODD.

There are no solutions to the equation because of this.

Therefore, integer values of a and b which satisfy the relationship = $\frac{a}{b}$ cannot be found.

Therefore $\sqrt{3}$ is irrational

Solution 5:

The following is a somewhat non-classical and less elementary proof.
Suppose that $\sqrt3\in\mathbb Q,$ and write $\alpha:=\sqrt3=p/q$ where $p$ and $q$ are relatively prime. Then we have $q\alpha=p,$ so in the ring $\mathbb Z[\alpha], \alpha$ divides $p,$ hence $\alpha^2\mid p^2,$ i.e. $3\mid p^2.$ So $p^2/3\in R:=\mathbb Z[\alpha]\cap\mathbb Q.$
In fact, this ring $R$ is nothing else but $\mathbb Z,$ by the lemma below, therefore $3$ divides $p^2,$ hence $p,$ in $\mathbb Z.$ So we conclude that $3$ divides $q\alpha$ as an integer; then $9\mid 3q^2.$ Since $p$ and $q$ are relatively prime, $9$ is prime to $q^2,$ thus $9\mid 3,$ a contradiction.

Lemma: $R:=\mathbb Z[\alpha]\cap\mathbb Q=\mathbb Z.$

For those who know some algebraic number theory already, this lemma needs no proof, but, for the sake of reference, I shall provide a brief proof of the above-used fact that $R=\mathbb Z,$ assuming known that $\mathbb Z[\alpha]$ is the integral closure of $\mathbb Z$ in $\mathbb Z(\alpha).$

Since $\mathbb Z[\alpha]$ is the integral closure of $\mathbb Z$ in $\mathbb Q(\alpha),$ $R$ consists of the elements in $\mathbb Q$ which are integral over $\mathbb Z,$ namely, of the elements $m/n\in\mathbb Q$ such that there exists a polynomial $f(x)=x^k+a_1x^{k-1}+\cdots+a_k$ with $a_i\in\mathbb Z, \forall i$ such that $f(m/n)=0.$ Then we might assume that $m$ and $n$ are relatively prime, and obtain the relation $m^k+a_1m^{k-1}n\cdots+a_kn^k=0,$ so every prime divisor of $n$ also divides $m,$ contradicting our hypothesis, unless $n=1,$ i.e. $m/n\in\mathbb Z.$ This proves that $R=\mathbb Z.$

P.S. This works for every square-free integer, after a slight modification.
Hope this may be of some interest, though not necessarily of any use. ;P