Show that 4√2 is an irrational number
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Let us assume that 4√2 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that 4√2 = a/b ∴ √2 = a/4b Since, a and b are integers, a/4b is a rational number and so √2 is a rational number. Alternate Proof: Let us assume that 4√2 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that Since, 32 divides a2, so 32 divides ‘a’ as well. So, we write a = 32c, where c is an integer. ∴ a2 = (32c)2 … [Squaring both the sides] ∴ 32b2 = 32 x 32c2 …[From(i)] ∴ b2 = 32c2 ∴ c2 = b2/32 Since, 32 divides b2, so 32 divides ‘b’. ∴ 32 divides both a and b. a and b have at least 32 as a common factor. But this contradicts the fact that a and b have no common factor other than 1. ∴ Our assumption that 4√2 is a rational number is wrong. ∴ 4√2 is an irrational number.
Answer:
Let us assume that is a rational number. ⇒ 4 2 = p q ,where p and q are the integers and q ≠ 0. Since, p,q and 4 are integers. So, p 4 q is a ratioanl number.
Let us assume that is a rational number. ⇒ 4 2 = p q ,where p and q are the integers and q ≠ 0. Since, p,q and 4 are integers. So, p 4 q is a ratioanl number.Step-by-step explanation:
Let us assume that is a rational number. ⇒ 4 2 = p q ,where p and q are the integers and q ≠ 0. Since, p,q and 4 are integers. So, p 4 q is a ratioanl number.Step-by-step explanation:make me brainleast