Hi guys, having problems solving this problem. Question is prove by contradiction that $\displaystyle \sqrt{3}$ is irrational. Any help would be highly appreciated.
Ok here is my attempt.
Suppose that $\displaystyle \sqrt{3}$ is rational. Therefore $\displaystyle \sqrt{3}$ = $\displaystyle \frac{m}{n}$ for some integers m, n with n $\displaystyle \neq$ 0.
We may suppose without loss of generality (wlog) that m, n have no common factor (i.e. are coprime). Now $\displaystyle m^2$ = $\displaystyle 3n^2$ and so $\displaystyle m$ is even. Put $\displaystyle m = 2k$. Then $\displaystyle 4k^2$ = $\displaystyle 2n^2$ and so $\displaystyle n^2$ = $\displaystyle 2k^2$
which implies that n is even. This contradicts the assumption that m, n are coprime.
Hence $\displaystyle \sqrt{3}$ must be irrational.
Is this correct?
Hello, jvignacio!
You've mixed in 2's, 4's, and even numbers.
Assume that $\displaystyle \sqrt{3}$ is rational.Prove by contradiction that $\displaystyle \sqrt{3}$ is irrational.
Then: .$\displaystyle \sqrt{3} \:=\:\frac{a}{b}$ for positive integers $\displaystyle a$ and $\displaystyle b$, where $\displaystyle a$ and $\displaystyle b$ are coprime. .[1]
Square the equation: .$\displaystyle \left(\sqrt{3}\right)^2 \:=\:\left(\frac{a}{b}\right)^2 \quad\Rightarrow\quad 3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2$
Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: .$\displaystyle a \,=\,3c$ .[2]
The equation becomes: .$\displaystyle 3b^2 \:=\:(3c)^2 \quad\Rightarrow\quad 3b^2 \:=\:9c^2 \quad\Rightarrow\quad b^2 \:=\:3c^2$
Since the right side is divisible by 3, the left side is also divisible by 3.
. . Hence: .$\displaystyle b \,=\,3d$ .[3]
But [2] and [3] say that $\displaystyle \text{GCD}\,(a,b) \,=\,3 $ . . . which contradicts [1].
. . $\displaystyle Q.E.D.$
That's because, if $\displaystyle a^2$ is divisible by 3, then so is $\displaystyle a$. In order for 3 to divide $\displaystyle a\cdot a$, one could try breaking 3 into two factors: $\displaystyle 3=pq$ so that $\displaystyle p\mid a $ and $\displaystyle q\mid a$, but $\displaystyle pq\not\;\mid a$ ($\displaystyle x\mid y$ means "$\displaystyle x$ divides $\displaystyle y$"). However, 3 is prime, and this is impossible.dont understand how you got to the a = 3c part?
Since $\displaystyle 3\mid a$, we have $\displaystyle 9\mid a^2$, and we can divide both sides of $\displaystyle 3b^2=a^2$ by 3.