I'm struggling with a textbook question thats asks to use proof by contradiction to show that (√10)/2 is an irrational number. I tried following a similar proof that the teacher did in class, but I ended up with some random values and I'm not quite sure what I did wrong/what steps I should take next. I'm also not sure if I'm doing proof by contradiction right? Any help would be appreciated.
My attempt:
1) (√10)/2 is rational
2) (√10)/2 = a/b
3) √10 = 2a/b
4) 10 =$(2a)^2$/$b^2$
5) 10 =$4a^2$/$b^2$
6) $10b^2$=$4a^2$
7) $5b^2$=$2a^2$
Prove $a^2$ is even → a is even
1) $a^2$ is even ∧ a is odd
2) $a^2$ is even
3) a is odd
4) a = 2k + 1
5) $a^2$ = $(2k+1)^2$
6) $a^2$ = $4k^2$ + 4k + 1 = 2($2k^2$ + 2k) + 1
7) $a^2$ = 2m + 1, given m = $2k^2$ + 2k
8) $a^2$ is odd
9) $a^2$ is even ∧ $a^2$ is odd
10) false
Therefore $a^2$ → a is even
$5b^2$ = $2a^2$
a = 2c from proof above
$5b^2$ = 2$(2c)^2$
$5b^2$ = 2($4c^2$)
$5b^2$ = $8c^2$
Start with assumption that $\frac{a}{b}$ is irreducible (if it's not, we can reduce it). Thus $a$ and $b$ cannot be both even. If $a$ is even and $b$ is odd, we have a contradiction as $5b^2$ is odd and $2a^2$ is even. If $b$ is even and $a$ is odd, we can write $b=2k$. Then we have that $20k^2=2a^2 \rightarrow 10k^2=a^2$ wich is a contradiction as $a^2$ is odd, $10k^2$ is even.
The last case (thanks, Bill Dubuque) is when $a$ and $b$ are odd and do not have common factors. Then $a=2k+1$, $b=2m+1$, $5b^2=20m^2+20m+5$ is odd, $2a^2=8k^2+8k+2$ is even.