Question

27. Prove that √5 is an irrational number



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Answer 1

Answer:

yes

Step-by-step explanation:

because root 5 = 5x5 =25 and it is rational no.

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Answer 2

verify by the method of contradiction

p:√5 is irrational number

answer:

In this method, we assume that the given statement is false. That is we assume that √5 is rational. This means that there exists positive integers a and b such that √7=a/b,

where a&b have no common factors. Squaring the equation, we get 7=a^2/b^2

→7 divides a.

Therefore, there exists an integer c such that a=7c. then a^2=49^2 and a^2=7b^2

Hence,7b^2=49c^2→b^2=7c^2→7 divides b.

but we have already shown that 7 divides a. This implies that 7 is common factor of both a and b which contradicts our earlier assumption √5 is rational is wrong.

Hence, the statement √5 is irrational number is true.