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| 1. |
Prove that √7 is an irrational |
| Answer» Ans. let us assume that {tex}\\sqrt 7{/tex}\xa0be rational.Then it must in the form of {tex}p \\over q {/tex}[q ≠ 0] [p and q are co-prime]{tex}\\sqrt7 = {p \\over q}{/tex}=>{tex}q \\sqrt 7 = p {/tex}Squaring on both sides{tex}=> 7q^2= p^2 {/tex} .............. (1)\xa0p2\xa0is divisible by 7p is divisible by 7p = 7c [c is a positive integer][squaring on both sides ]p2\xa0= 49 c2\xa0..............\xa0(2)Subsitute p2\xa0in eq\xa0(1) we get=> 7q2\xa0= 49c2\xa0=>q2 = 7c2=> q is divisble by 7 thus q and p have a common factor 7.There is a contradiction as our assumsion p and q are co primebut it has a common factor.So that {tex}\\sqrt 7{/tex}\xa0is an irrational. | |