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| 1. |
Prove that √7 is irrational no. |
| Answer» Let us assume that {tex}7^1/2{/tex}\xa0is a rational number. So it can be written in the form of a/b.{tex}a/b = 7 ^1/2{/tex}Squaring both sides, we get\xa0{tex}a^2/ b^2 = 7{/tex}{tex}a^2 = 7 b^2{/tex}\xa0......(1)This means that\xa0{tex}a^2{/tex}\xa0is divisible by 7. This implies that a is also divisible by 7.let a = 7c\xa0squaring both sides , we get\xa0{tex}a^2 = (7c)^2{/tex}\xa0or\xa0{tex}7b^2 = 49 c ^2{/tex}or\xa0{tex}b^2 = 7c^2{/tex}......(2)From eq(1) and eq(2), it is clear that both a and b are divisible by 7. . An irrational number can not common factor.This shows that our assumption is wrong.\xa0So\xa0{tex}7^1/2 {/tex}\xa0is an irrational number.\xa0 | |