Question

p.t 7/3 root5 is an irrational no

Answer 1

step by step i need d answer cn any1 plss help me .................. :(

Answer 2

is this your question?

Answer 3

I assume that "p.t" means "prove that"?

Answer 4

yaaaaaaaaaa @ParthKohli and @amorfide

Answer 5

I don't know the proof sorry

Answer 6

ok @ParthKohli cn u do it

Answer 7

@Hero cn u help mee

Answer 8

Hmm. Assume that your number is rational.\[{7 \over 3\sqrt5} = {p \over q} \,\wedge \gcd (p,q) = 1\]

Answer 9

\[{49 \over45 } = {p^2 \over q^2}\]\[49q^2 =45p^2 \]This doesn't seem like the conventional proof. :(

Answer 10

is it correct

Answer 11

No, I need to find a contradiction which I am unable to do.

Answer 12

uff will u try 2 doo ma frnddd

Answer 13

7/3sqrt5 = p/q -> sqrt 5 = 7q/3p and square ->

5=49q^2/9p^2 so p^2/q^2 = 49/45 but no integer squares to 45

Answer 14

Oh, so that was the contradiction!