Question
$\sqrt{2}$ is irrational.
✓
Answer
Let us assume, to the contrary, that $\sqrt{2}$ is rational.
So, we can find integers $r$ and $s(\neq 0)$ such that $\sqrt{2}=\frac{r}{s}$.
Suppose $r$ and $s$ have a common factor other than 1 . Then, we divide by the common factor to get $\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are coprime.
So, $b \sqrt{2}=a$.
Squaring on both sides and rearranging, we get $2 b^2=a^2$. Therefore, 2 divides $a^2$.
Now, by Theorem 1.3, it follows that 2 divides $a$.
So, we can write $a=2 c$ for some integer $c$.
Substituting for $a$, we get $2 b^2=4 c^2$, that is, $b^2=2 c^2$.
This means that 2 divides $b^2$, and so 2 divides $b$ (again using Theorem 1.3 with $p=2$ ).
Therefore, $a$ and $b$ have at least 2 as a common factor.
But this contradicts the fact that $a$ and $b$ have no common factors other than 1 .
This contradiction has arisen because of our incorrect assumption that $\sqrt{2}$ is rational.
So, we conclude that $\sqrt{2}$ is irrational.
So, we can find integers $r$ and $s(\neq 0)$ such that $\sqrt{2}=\frac{r}{s}$.
Suppose $r$ and $s$ have a common factor other than 1 . Then, we divide by the common factor to get $\sqrt{2}=\frac{a}{b}$, where $a$ and $b$ are coprime.
So, $b \sqrt{2}=a$.
Squaring on both sides and rearranging, we get $2 b^2=a^2$. Therefore, 2 divides $a^2$.
Now, by Theorem 1.3, it follows that 2 divides $a$.
So, we can write $a=2 c$ for some integer $c$.
Substituting for $a$, we get $2 b^2=4 c^2$, that is, $b^2=2 c^2$.
This means that 2 divides $b^2$, and so 2 divides $b$ (again using Theorem 1.3 with $p=2$ ).
Therefore, $a$ and $b$ have at least 2 as a common factor.
But this contradicts the fact that $a$ and $b$ have no common factors other than 1 .
This contradiction has arisen because of our incorrect assumption that $\sqrt{2}$ is rational.
So, we conclude that $\sqrt{2}$ is irrational.
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