Question

prove that 6 + root 2 is an irrational number​

Answer 1

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Let 6+ √2 is a rational number then we get that a and b two co-prime integers.

Such that 6+ √2=a/b where b not equal

√2 = a/b - 6

√2 =(6-a/b)

Since a and b are two integers.

Therefore (6-a/b)is a rational number and So root also is a rational number.

But it is contradiction to fact root 2 =(6-a/b) is rational number.

So we that 6+ √2 is an irrational number.

Answer 2

Answer:

let √2be rational num

√2=p/q. where p and q are coprimes q≠0

(√2)^2=p^2/q^2

2q^2=p^2

as p^2 must be divisible by 2 there fore p must be divisible by 2

p=2c where c is constant

2q^2=2c^2

similarly as p,q must be also divisible by 2

as we discussed before that they are co prime

this is contradiction to our original assumption

=>√2is irrational no.

therefore when we are adding 6+√2

this gives us an irrational no

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