Question

plz tell me the answer of this question. prove that under root of 7 is irrational​

Answer 1

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⬇let us assume that √7 be rational.⬇

⬇then it must in the form of p / q.⬇

➡As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.⬅

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➡ √7 = p / q

➡ √7 x q = p

◽squaring on both sides ⭐⭐⭐⬇⬇

➡ 7q² = p² ------1.

▶ p is divisible by 7

◽p = 7c [c is a positive integer] [squaring on both sides ]⬇⬇⬇

▶ p²= 49c²

➡➡substitute p² in eq.(1) we get⬅⬅

➡ 7q² = 49 c²

➡ q² = 7c²

➡ q is divisible by 7

⭐thus q and p have a common factor 7. there is a contradiction to our assumption as our assumesion p & q are co prime but it has a common factor.⭐

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so that √7 is an irrational....

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hope it helps you..⭐⭐⭐

plz Mark as Brilliant⭐⭐⭐⭐⭐

Answer 2

Answer:

√7=a/b ( here a and b are co prime means they have only 1 as common factor. ... Here we find 7 is common which divide both a and b but this is contradiction because a and b are co prime they don't have common factor other than 1. So for our assumption is wrong. Hence √7 is irrational.