prove that 2+ root5 is irrational number

Question

anonymous · Answer

WELCOME TO OPEN STUDY and bc it haas many numbers after the decimal point in a random order here is the proof 2+root5=4.2360679774997896964091736687313

geerky42 · Answer

Here's the better structured proof:

We will prove it using proof by contradiction. Suppose $2+\sqrt5$ is rational $r$, so that $$2+\sqrt5=r$$Squaring both sides, we have $4+4\sqrt5+5=r^2$, which means that $\sqrt{5}=\dfrac{r^2-9}{4}$

Since the set of rational numbers is closed under division, multiplication, and subtraction, $\dfrac{r^2-9}{4}$ is rational. However it is known that $\sqrt5$ is irrational. We have a contradiction.

Therefore $2+\sqrt5$ is irrational.  $\blacksquare$

anonymous · Answer

Another way to see $2+\sqrt{5}$ is irrational is as follows. If $x=2+\sqrt{5}$, then: $$x=2+\sqrt{5}\Longrightarrow x-2=\sqrt{5}\Longrightarrow \left(x-2\right)^2=5$$ $$\Longrightarrow x^2-4x+4=5\Longrightarrow x^2-4x-1=0.$$Thus $2+\sqrt{5}$ is a root of the polynomial $x^2-4x-1$.  If $2+\sqrt{5}$ were rational, it would be a rational root of the the above polynomial. However, using the rational roots theorem, we see the only possible rational roots are $\pm 1$, neither of which are actually roots. Thus $2+\sqrt{5}$ cannot be rational.