Question

prove that
2root3+root5 is an irrational

Answer 1

I hope its help u ..............

Answer 2

Hey..!!!

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If root3 + root5 is rational therefore:

3√+5√=a/b

and a/b is fraction, meaning it can be simplified form. (ie: 6/10 = 3/5) therefore either a or b or both must be odd. Because if it was even then you could simplify it by cancelling out a 2. Then this implies:

3√+5√=a/b

(3√+5√)2=a2/b2

(8+215√)=a2/b2

2(4+15√)=a2/b2

2∗b2(4+15√)=a2

This implies that a^2 is even, since a^2 = 2*b^2(4+root15) Which also implies that
(a) itself is also even (if a was odd, then a^2 = odd*odd which is also odd, but we know a^2 is even, so it must be that
(a) is even). since a = even, we know that it can be written in the form a =2n now plug this in for (a)

2×b2(4+15√)=a2

2×b2(4+15√)=4n2

b2(4+15√)=2n2

but wait! this says that if (a) is even, then (b) is also even! In the beginning we assumed that root3+root5 is rational, then either a or b or both must be odd. and that is a contradiction because it is saying that if: root3 + root5 = rational then the sum will give you a number such that (2n/2k) =/= (n/k) so it must be irrational.

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I Hope it's help you...!!!