Prove $1+2\sqrt3$ is not a rational number

Solution 1:

Say $1 + 2 \sqrt{3} $ is rational by contradiction. So we have

$$ 1 + 2 \sqrt{3} = \frac{p}{q} $$

for $p,q \in \mathbb{Z}$. Notice we can move $1$ to the other side to obtain

$$ 2 \sqrt{3} = \frac{p}{q} - 1 = \frac{p-q}{q}$$

Now, divide by $2$ to obtain

$$ \sqrt{3} = \frac{p-q}{2q} $$

But this is a rational number. Hence $\sqrt{3} \in \mathbb{Q}$. And we have reached a contradiction since $\sqrt{3}$ is not rational.

Solution 2:

From the question you made I assume you are beginning your studies in mathematics, so additionally to Willie Rosario's answer you will need to prove that $\sqrt{3}$ is irrational as well, and that's how you do:

Assume $\sqrt{3} \in \mathbb{Q}$. Then, $\sqrt{3}=p/q$, where $\gcd\{p,q\}=1$ (indeed, if this was not true, you could just simplify the fraction). This means that if $$ p=p_1p_2\ldots p_n, \quad q=q_1q_2\ldots q_m $$ are the decompositions of $p,q$ on prime factors, then $p_j\neq q_k$ for every $j,k$, since $\gcd\{p,q\}=1$. But now, $$ 3=\frac{p^2}{q^2} $$ means that $q^2$ divides $p^2$, which is impossible since $$ p^2=p_1^2p_2^2\ldots p_n^2, \quad q^2=q_1^2q_2^2\ldots q_m^2, $$ and we know that $p_j\neq q_k$ for all $j\neq k$ (the fact that $p^2$ divides $q^2$ implies that there exist $j,k$ such that $p_j=q_k$). So, we arrive at a contradiction, and we conclude that $\sqrt{3}\notin \mathbb{Q}$.

Solution 3:

$x = 1 + 2\sqrt{3} \Rightarrow (x-1)^2 = 12 \Rightarrow x^2-2x-11 = 0$. Using the rational root test, there is no rational root, hence the conclusion...

Solution 4:

If it was rational then $1+2 \sqrt{3} = { p \over q}$ and so $\sqrt{3} = {1 \over 2} ({ p \over q} -1)$, which would mean that $\sqrt{3}$ is rational.

Solution 5:

$\,r = 1+2\sqrt{3}\,\Rightarrow\, (r-1)^2 = 12.\,$ By the Rational Root Test, if it $\,r\,$ is rational then it is an integer. Therefore $\,n = r-1\,$ is an integer and $\,n^2 = 12,\,$ contra $\, 3^2 < 12 < 4^2,\,$ and $\,x^2\,$ is increasing (or contra $\,12\,$ is not a square mod $5)$