irrational numbers

I will give some examples about irrational numbers. Irrational numbers are numbers that cannot be expressed by dividing any integer $p$ by any integer $q$. According to this definition [Def-2], $p$ and $q$ are relative prime (integer) numbers.

In order to prove that a target number is irrational,  we use a proof by contradiction. Let the target number be expressed by $p/q$. Then, if any contradiction will occur, it follows that the number is not a rational number. Here are some examples.

If $\sqrt{2}$ is rational, we are able to assume $\sqrt{2}=p/q$. Since $p^2=2q^2$, $p^2$ is an even number. In this case, $p$ must be an even number, too. If $p$ is an odd number, $p^2=(2m+1)^2=2(2m^2+2m)+1$, $p^2$ is also an odd number. Therefore, $p$ can not be an odd number.

Thus, as we are able to put $p=2m$, we get $2q^2=4m^2$. Both sides of this equation can be divided by 2. However, if $p$ and $q$ are relative prime numbers, $p^2$ and $q^2$ must be relative prime numbers. it is inconsistent with [Def-2] in which $p$ and $q$ are relative prime numbers. Therefore, $\sqrt{2}$ is an irrational number.

According to this method, we can prove that $\sqrt{3}$ is irrational. It is important that when $p^2=3q^2$, $p$ must be a multiple of 3. Suppose $p=3m+1$. Then $p^2=(3m+1)^2=3(3m^2+2m)+1$. Therefore $p^2$ is not a multiple of 3. If $p=3m+2$, also then $p^2=(3m+2)^2=3(3m^2+4m+1)+1$. $p^2$ is not a multiple of 3, either. Hence, we get $p=3m$. Similarly, since it is a contradiction of [Def-2], Hence $\sqrt{3}$ is an irrational number.

It is even more difficult to show that $\pi$ is an irrational number. However, it has already been successfully done by contradiction. In mathematics, since rational numbers are at most countable and irrational numbers are uncountable, irrational numbers take a greater part in real numbers. We are not able to figure out irrational numbers. For example, we do not know whether $\pi+e$ is an irrational number, or not.