prove that rt2 + rt3 is irrational

fishpaste

not sure how to start, can't think of any arguments which are going to result in anytihng.

Anybody? :smile:

(rep available!)

Reply 1

JamesF

Assume it is rational for a contradiction.

Reply 2

JamesF

mockel

Isn't any irrational number + another irrational number always an irrational number?

(2 + rt2) + (2 - rt2) = 4

Reply 3

m:)ckel

JamesF

(2 + rt2) + (2 - rt2) = 4

wow, i didn't think of that :redface:

just use your method then...

Reply 4

fishpaste

JamesF

Assume it is rational for a contradiction.

Hm okey

rt2 + rt3 = p/q st. (p,q) = 1

(5 + 2rt6)q^2 = p^2

RHS is an integer

RTP: (5 + 2rt6)q^2 is not an integer

q^2 is an integer, LHS = (rational + irrational)*rational = irrational

therefore contradiction.

I'll try and make it a bit more rigorous. But I think this is a fine argument? THANKS :smile:

Reply 5

fishpaste

how can I show that a rational + irrational is irrational? decimal expansions?

Reply 6

JamesF

fishpaste

Thats fine. The only thing i would do, and its just being picky, is i would expand it slightly, since "LHS = (rational + irrational)*rational = irrational" is not *that* obvious
so, i would add a few more lines to make it crystal clear;
(5 + 2rt6)q^2 = p^2
2q^2*(rt6) = p^2 - 5q^2
rt(6) = (p^2 - 5q^2)/2q^2

RHS is rational, which is a contradiction because rt6 is not rational.

Reply 7

fishpaste

how can I show that a rational + irrational is irrational? decimal expansions?

sorted this, note to self, try just a bit harder before asking for help

Reply 8

JamesF

fishpaste

how can I show that a rational + irrational is irrational? decimal expansions?

Exactly the same way, group the rational terms on one side and the irrational ones on the other - contradiction.

Reply 9

Gauss

assume rt2 is rational

then u can say rt2 = p/q where p,q are integers, with highest common factor 1

hence 2 = p^2/q^2
=> q^2*2 = p^2
=> p^2 is even
=> p is even

now u can express p as an even number:
p = 2k for some integer k
=> p^2 = 4k^2
=>2*q^2 = 4k^2
=>q^2 = 2k^2
=>q^2 is even
=> q is even

hence if q is even and p is even then the highest common factor of p and q is not 1, hence rt2 cannot be expressed as p/q
=> rt2 is irrational.

u can use the same method of proof for the second part, however in the middle u have to prove that if 3 divides n^2 then 3 divides n

thats simple too, just consider the remainders when 3 is divided by n

hope that helped