prove that if square root of 6 is irrational then square root of 3 plus square root of 2 is also irrational?

Question

if sqrt6 is irrational then sqrt3+sqrt2 is also irrational prove it.. another is by using law of arthmetic prove that asqr+bsqr+csqr=< (a.b+a.c+b.c+a.c)...thanks

Answer 1

It suffices to prove the contrapositive: if √2 + √3 is rational, then √6 is rational.

Suppose √2 + √3 is rational, say,

√2 + √3 = m/n

where m and n are positive integers.

Then

(√2 + √3)² = m²/n²

Expand the left side, and you will be able to express √6 as a rational number.

I know of a way to prove a² + b² + c² ≤ 3(ab + ac + bc) using calculus, but I suspect that's not what you're looking for.

Answer 2

assume √3 is not irrational so integers a and b can be found that √3= a/b and a and b are relatively prime (gcd(a,b)=1) and b≠1 √3= a/b => 3 = a^2 / b^2 => a^2 = 3 * b^2 By this point we conclude that a^2 is dividable by b^2 since we have gcd(a,b)=1 we conclude that a is dividable by b as well.(It so happens to be true that, for any integers m and n, m^2 | n^2 implies that m | n. The proof of this is rather more complex than can be gotten into now) formal notation: b^2 | a^2 and gcd(a,b)=1 => b | a so since gcd(a,b) = 1 and b | a it implies that b=1; However, above we assumed that b ≠ 1 so there is a contradiction. and our assumption is wrong. so √3 is irrational.

Answer 3

the square root of 6 is 2.4494897427831780981972840747059 and it keeps going...so its irrational and the square root of 3 is 1.7320508075688772935274463415059 and the square root of 2 is 1.4142135623730950488016887242097 and the square root of 3 plus the square root of 2 equals 3.1462643699419723423291350657149... which continues for a while stating that it is also irrational...so there you go