The first part is correct, but you could clarify it a little bit:
Since $p$ and $q$ are irrational, they are not integers
Since $p$ and $q$ are not integers, $p^2$ and $q^2$ are not perfect squares
Since in addition to that $p^2$ and $q^2$ are relatively prime, $p^2q^2$ is not a perfect square
Since $p^2q^2$ is not a perfect square, $\sqrt{p^2q^2}=pq$ is irrational
The second part is somewhat obscure, but you could simply use the following argument instead:
Since $pq$ is irrational, it is not a multiple of two integers
Since $pq$ is not a multiple of two integers, it is not a perfect square
Since $pq$ is not a perfect square, $\sqrt{pq}$ is irrational
....
Therefore, the prime factorization of 2 contains an odd exponent so 2 is not a perfect square.
Right. But that only goes to prove that if $\sqrt{2}^2 = 2$, then $\sqrt{2}$ is not an integer. We need to prove it is not a rational number.
And if you want to use your theorem:
If $\sqrt{2} = \frac ab$ where $a$ and $b$ are integers, means that $2 = \frac {a^2}{b^2}$ which means that $2b^2 = a^2$.
Now $b^2$ is perfect square. So it's prime factorization has only even powers. So the power of $2$ that factors into $b^2$ is even. (If $b$ is odd then $0$ is an even number). So the power that divide into $2b^2$ will be $1$ higher.
So the power of $2$ that divides into $2b^2$ is odd. So $2b^2$ is not a perfect square.
But $2b^2 = a^2$ so it is a perfect square.
That's a contradiction. So our assumption that there are integers $a,b$ so that $\sqrt{2}=\frac ab$ can't be true. So $\sqrt{2}$ is not rational.
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Or maybe our theorem that A number is a perfect square if and only if its prime factorization contains only even exponents is incorrect.
It is correct and we can use it but we do have to prove it before we take it for granted.
..... PS .....
Note: By your theorem for any prime $p$ then and integer $n$ we can see $pn^2$ is not a perfect square and indeed of any non-square $j$ then $jn^2$ can not be a perfect square.
..... PPS......
Throughout all of this we assume that there is such a thing as "$\sqrt{2}$". There might not be. After all if we are dealing with real numbers there is no such thing as $\sqrt{-1}$. And if we are dealing with complex numbers there is no such thing as $\sqrt{\text{Babar, the elephant}}$.
It doesn't matter now, but later it will.
Our proof isn't really that "$\sqrt{2}$ is irrational" but that "if we define $\sqrt{2}$ as the number $r$ so that $r^2 = 2$ then either $\sqrt{2}$ does not exist at all or that $\sqrt{2}$ is irrational", or to put in more legitimately "There is no rational number $r$ so that $r^2 = 2$".
To prove that there is a real number $r$ so that $r^2 = 2$ is a whole other kettle of fish.
Best Answer
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.