Prove irrational

Question

Prove that $\dfrac{1+\sqrt{5}}{2}$ is irrational.

a.)Without assuming $\sqrt{5}$ is irrational.

b.)With assuming $\sqrt{5}$ is irrational.

For a.) would I just do a proof by contradiction. Assuming that the entire expression is rational?

score 2 · Answer 1 · edited Oct 29 '17 at 18:31

2

Suppose $${1+\sqrt{5}\over 2} = q\in \mathbb{Q}\;\;\; \Longrightarrow \;\;\;\sqrt{5} =2q-1$$

So if we already know that $\sqrt{5}$ is irrational then we are finished since $2q-1$ is rational. Now suppose we don't know that. Then $2q-1 ={a\over b}$ for some relatively prime integers $a,b$. So $5 ={a^2\over b^2}$ and thus $a^2=5b^2$.

So $5|a^2$ so $5|a$ and we can write $a=5c$. Now we have $25c^2 = 5b^2$ and so $5|b^2$ and we have a contradiction since we assumed that $a,b$ are relatively prime.

edited Oct 29 '17 at 18:31

Frentos

3,041

answered Oct 29 '17 at 18:04

nonuser

90,026

how are you getting a=5c? what is c – tskgreen Oct 29 '17 at 18:15
oh never mind i see. – tskgreen Oct 29 '17 at 18:16
@JohnWatson yes, very interesting. :-) – S. Y Oct 29 '17 at 18:18

score 0 · Answer 2 · answered Oct 29 '17 at 18:11

0

For a) consider the polynomial $x^2-x-1=0$ then $x=\frac{1+\sqrt{5}}{2}$ is a root of the polynomial, now can you prove that there are no rational number $x$ such that $x^2-x-1=0$ ?

You can prove it by contradiction if you want.

answered Oct 29 '17 at 18:11

kingW3

13,496

score 0 · Answer 3 · answered Oct 29 '17 at 18:55

Actually, it's for b) that a proof by contradiction would work very well:

If we assume that $\dfrac{1+\sqrt{5}}{2}$ is rational, then $\dfrac{1+\sqrt{5}}{2} = \frac{m}{n}$ for some integers $m$ and $n$.

Hence, $1 + \sqrt{5} = \frac{2m}{n}$, and therefore:

$$\sqrt{5} = \frac{2m}{n} -1 = \frac{2m-n}{n}$$

Since $m$ and $n$ are integers, $2m-n$ and $n$ are as well, which means that $\sqrt{5}$ is rational, which contradicts the assumption that it isn't.

Prove irrational

3 Answers3