Prove that 3+2√5 irrational

Let's assume, for the sake of contradiction, that $3 + 2\sqrt{5}$ is a rational number.
If it is rational, then it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are integers, $b \neq 0$, and $\gcd(a,b) = 1$ (meaning the fraction is in its simplest form).

Subtract 3 from both sides of the equation:

To combine the terms on the right side, find a common denominator:

Now, divide both sides by 2:

Since $a$ and $b$ are integers, and $b \neq 0$:

• $a - 3b$ is an integer (because the difference and product of integers are integers).
• $2b$ is an integer (because the product of integers is an integer).
• Since $b \neq 0$, $2b \neq 0$. Therefore, the expression $\frac{a - 3b}{2b}$ is a ratio of two integers where the denominator is non-zero. By definition, this means $\frac{a - 3b}{2b}$ is a rational number.

It is a known mathematical fact that $\sqrt{5}$ is an irrational number. (This can be proven separately, for example, by assuming $\sqrt{5} = \frac{p}{q}$ and showing $p^2 = 5q^2$ implies both $p$ and $q$ are multiples of 5, leading to a contradiction of $\gcd(p,q)=1$).

From Step 2, we derived that $\sqrt{5} = \frac{a - 3b}{2b}$.
From Step 3, we established that $\frac{a - 3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However, from Step 4, we know that $\sqrt{5}$ is an irrational number.
This creates a contradiction: an irrational number cannot be equal to a rational number.

Our initial assumption that $3 + 2\sqrt{5}$ is rational must be false.
Therefore, $3 + 2\sqrt{5}$ is an irrational number.