prove. (3√2+√5) Is an irrational. number. and. also check. whether (3√2+√5)(3√2-√5) is irrational. or rational.

1
by prabhushettisan

2014-10-18T05:25:25+05:30

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Let us assume 3√2 + √5 = a rational number = p / q  where p and q are integers and q is not 0.   Let us assume that p / q is in reduced form and have no common factors and are prime to each other.

(3 √2 + √5 )² = p² / q²
18 + 5 + 6 √10 = p² / q²
√10 = (p²-23 q²) / 6 q²  =  m/n where m and n are integers and n≠ 0

10 = m² / n²       or,     m * m = 2 * 5 * n * n

Since m and n are prime to each other, m must have a factor of 2 and a factor of 5 also.  Let   m = 2 * 5* k.

2 * 5 * k * 2 * 5 * k = 2 * 5 * n * n
2 * 5 * k * k = n * n
Hence n must have as a factor 2 and 5 also.  So, n = 2 * 5 * h.  We started with p/q where they are co-prime and derived that they have common factors 2 and 5.   Our assumption that the given number is a rational number , must be wrong.

Hence, 3 √2 + √5  is irrational.

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we know that (a +b)(a - b) = a² - b²
(3√2 + √5) (3√2 - √5) = (3√2)² - (√5)²  = 9*2 - 5 = 13

So the product of the two irrational numbers is rational and in fact, is a positive integer.
Sir here how u got √10=(p2-23q2)/6q2 and u wrote q2 two times and we can't. Separate 6√10