Prove that there is no positive rational number a such that $a^2 = 3$. I solved a similar question $a^4 = 2$ but I didn't find it hard. This one is taking a lot of my time and I still don't understand how to solve this.
Here's what I have tried so far.
edit:
\begin{align}
a^2 &= 3 \\
a &= \left(\frac mn\right) \\ 
a^2 &= \left(\frac mn\right)^2 = 3 \\
m^2 &= 3n^2 \\
m &= 3p \\
m^2 &= (3p)^2 \\
9p^2 &= 3n^2
\end{align}
In another similar question $a^3 = 2$, I had to divide a common number from each side to finally solve the problem. But with this problem, I don't know how to continue.
 A: Alright, make the straightforward assumption that $a= \frac mn$ with $m,n$ in lowest terms and they are integers. Thus, we get $(\frac mn)^2 = 3$, which soon becomes $m^2 = 3n^2$. Thus, $m^2$ is a multiple of $3$.
However, for $m^2$ to be a multiple of $3$, it needs to have an even number of factors of $3$. If it had an odd number of factors of $3$, you wouldn't get an integer for $m$ when you take the square root of $m^2$ because squares come in factors of even powers like $3^2$ or $2^2$. Thus, you must have an even number of $3$s in $m^2$, for it to be divisible by $3$. Now, when we take the square root of $m^2$ to get $m$, half of the $3$s disappear, but not all of them, so $m$ is divisible by $3$. Since $m$ is divisible by $3$, it can be expressed as $m=3p$. Thus, we get $(3p)^2 = 3n^2 \implies 3p^2 = n^2 $. Using the same logic as before, we can show that $n$ is divisible by $3$. This means that both $m,n$ are divisible by $3$, contradicting our initial assumption that $\frac mn$ is in lowest terms and shouldn't have any common factors besides $1$. Thus, there is no such $a = \frac mn$ with $m,n$ in lowest terms and integers such that $a^2=3$.
Here is a bonus easy proof. Rearrange $a^2=3$ to get $a^2-3=0$. By the rational root theorem, all the POSSIBLE rational roots of this polynomial are $a=-1,1,-3,3$. Let's see if any of these $a$ make $a^2=3$. Testing all these $a$, we see that none of these work, thus there is no RATIONAL $a$ such that $a^2=3$.
A: The classic method known as "infinite descent". Assume from where you had : $m^2 = 3n^2 \implies m = 3k \implies n^2 = 3k^2 \implies n = 3p$, and it keeps going indefinitely with new pairs $(k,p)$ with $k < m, p < n$ and both are multiple of $3$ which is impossible. Thus there is no $(m,n)$ which satisfies the equation. So $\sqrt{3}$ must be irrational.
A: From
$$ m^2 = 3n^2  \text{,}  $$ we know $3$ divides $m$, so $m = 3 m_1$
$$  9 m_1^2 = 3 n^2  $$
$$  3 m_1^2 = n^2  $$
This tells us $3$ divides $n$.  We have found that $3$ is a common factor of $m$ and $n$.
You failed to introduce $m$ or $n$ in your derivation and you are missing any words that would make this common divisor lead to a contradiction.  Perhaps you believe that a proper proof is just a barrage of equations.  It is not.  You need to introduce new symbols and explain how they relate.  For instance, a reader is right to ask "Why did you set $a = m/n$?  What is $m$?  What is $n$?  Are there any conditions among $m$ and $n$?"  Your demonstration is not a proof -- it is missing all the words that explain what you are doing and why.
