Number System - Prove irrationality of sqrt(2) and sqrt(3) using contradiction-based reasoning

Prove that $\sqrt{2}$ is an irrational number.

1. Assume to the contrary that $\sqrt{2}$ is rational.
2. Then $\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $a, b$ are coprime (have no common factors other than $1$).
3. Squaring both sides: $2 = \frac{a^2}{b^2} \implies a^2 = 2b^2$
4. This means $2$ divides $a^2$. By the theorem, if $2$ divides $a^2$, then $2$ divides $a$.
5. Let $a = 2c$ for some integer $c$. Substituting this in $a^2 = 2b^2$: $(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2$
6. This means $2$ divides $b^2$, so $2$ divides $b$.
7. Therefore, $a$ and $b$ have at least $2$ as a common factor. This contradicts the assumption that $a$ and $b$ are coprime.
8. Hence, $\sqrt{2}$ is irrational.

The proof uses the property that if a prime divides the square of an integer, it must divide the integer itself. Finding a common factor of $2$ for both $a$ and $b$ contradicts our simplest-form assumption.

Prove that $\sqrt{3}$ is an irrational number.

1. Assume $\sqrt{3}$ is rational. Let $\sqrt{3} = \frac{a}{b}$ where $a, b$ are coprime integers and $b \neq 0$.
2. Square both sides: $3 = \frac{a^2}{b^2} \implies a^2 = 3b^2$
3. Since $3$ divides $a^2$, $3$ must divide $a$.
4. Let $a = 3k$. Substitute into the equation: $(3k)^2 = 3b^2 \implies 9k^2 = 3b^2 \implies b^2 = 3k^2$
5. Since $3$ divides $b^2$, $3$ must divide $b$.
6. Both $a$ and $b$ have $3$ as a common factor, contradicting that they are coprime.
7. Thus, $\sqrt{3}$ is irrational.

Similar to the $\sqrt{2}$ proof, we demonstrate that both the numerator and denominator share a common factor of $3$, which violates the definition of a rational number in its simplest form.