Can anybody help on proving sq. root of 3 is not rational?
Indirect Method.
thank you so much.
If sqrt(3) was rational, then by definition it could be written as a/b, where a and and b are both integers. To prove indirectly that sqrt(3) is irrational, we must prove that a and b can not exist.
Suppose: sqrt(3) = a/b, a and b are integers (it is rational), and a/b is in lowest terms
1) Squaring both sides yields: 3 = a^2/b^2
2) Multiplying both sides by b^3 yields: 3b^2 = a^2
- Suppose b is even. Then b^2 is even and 3b^2 is also even. This means that a^2 and a must also be even. If both a and b are even numbers, then the ratio a/b can be simplified by cancelling a common factor of two, and contradicts our assumption that a/b is in lowest terms. For the sake of this proof, we must then assume that a and b are therefore both odd and that a/b is written in simplest terms.
3) Since a and b are odd, we can write the equations:
a = 2x+1
b = 2y+1
Where x and y are integers, thus restricting a and b to be any odd integer.
Substituting these into 3b^2 = a^2, we get:
3(2y+1)^2 = (2x+1)^2 (simplify)
3(4y^2+4y+1) = 4x^2+4x+1 (distribute)
12y^2+12y+3 = 4x^2+4x+1 (subtract 1)
12y^2+12y+2 = 4x^2+4x (factor)
2(6y^2+6y+1) = 2(2x^2+2x) (divide by 2)
6y^2+6y+1 = 2x^2+2x (factor right side)
6y^2+6y+1 = 2(x^2+1)
Now, for any choice of "y", the left hand side will be odd. For any choice of "x", the right hand side will be even. Thus, there exist no solutions to this equation, and proof that sqrt(3) is irrational is evident by contradiction.
Let me know if you have any questions!
That seems to be a lot of work.
I would start as you did, supposing $\displaystyle \frac{a}{b}=\sqrt{3}$, with $\displaystyle \gcd (a,b)=1$. Then, squaring both sides, $\displaystyle \frac{a^2}{b^2}=3\Rightarrow a^2=3b^2$.
So $\displaystyle 3\mid a^2$. Since 3 is prime, this implies $\displaystyle 3\mid a$, from which we conclude that we actually have $\displaystyle 9\mid a^2$.
Hence $\displaystyle 9\mid 3b^2$ as well. It follows that $\displaystyle 3\mid b^2$, and finally, $\displaystyle 3\mid b$.
Thus, $\displaystyle 3\mid \gcd (a,b)$, or $\displaystyle 3\mid 1$. This is, of course, a contradiction (well, in $\displaystyle \mathbb{Z}$).
Instead of what you do next, here is another way (I think nicer and a bit shorter) :
$\displaystyle a^2 = 3b^2$ so $\displaystyle b|a^2$. Since $\displaystyle a\wedge b=1$, $\displaystyle a^2\wedge b = 1$ as well. Euclid's lemma give us so $\displaystyle b | 1$, that is $\displaystyle b=1$.
As well, $\displaystyle a^2 = 3b^2$ says that $\displaystyle a|3b^2$ and so $\displaystyle a|3$ (with Euclid's lemma again). So $\displaystyle a\in\{1,3\}$. Both gives us a contradiction ($\displaystyle 1=3$ or $\displaystyle 9=3$, which is the same by the way).
I like it, never thought of doing it that way.
Of course... can't we just say that ANY equation of the form $\displaystyle a^2=nb^2$, with $\displaystyle n\in \mathbb{N}\setminus \{0\}$ not a square, is a contradiction simply by observing that this must give us some prime occurring to BOTH an odd and even power in the same factorization, contrary to the fact that $\displaystyle \mathbb{Z}$ is a unique factorization domain? Err... contrary to the Fundamental Theorem of Arithmetic, I suppose it is?
Yes, it's a way to do it, clearly.
What's bothering me it's the use of the factorial structure of the ring $\displaystyle \mathbb Z$, when our two proofs above seem to only use its euclidian structure...
(I said 'seem' because one might be have to use the factorial structure for something like $\displaystyle 3|a^2 \Rightarrow 3|a$ [in your proof] or $\displaystyle a\wedge b = 1 \Rightarrow a^2 \wedge b = 1$ [in mine]. I will check that and then tell if my remark is relevant or not.)
Why not just prove this in general?
If p is a positive integer which is not a square then $\displaystyle \sqrt{p}$ is irrational.
The proof by contradiction is easy. If $\displaystyle K$ is the least positive integer having the property that $\displaystyle K\sqrt{p}\in\mathbb{Z}^+$ then $\displaystyle K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor $ is smaller positive integer also having that property.
Another method:
Let us denoting by $\displaystyle P(x)$ the polynomial $\displaystyle x^2-3$,
$\displaystyle P(x)=x^2-3$, the only possible solutions of $\displaystyle P(x)$ over $\displaystyle \mathbb{Q}$ are: $\displaystyle \pm{1},\pm{3}$, and non of them is the solution of $\displaystyle P(x)$.