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Prove that √3+√4 is an irrational number |
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Answer» Lets assume that : √3 + √4 is rational. √3 + √4 = r , where r is rational Squaring both sides , we get [√3 + √4 ]² = r² 3 + 2√12 + 4 = r² 7 + 2√12 = r² 2√12 = r² - 6 √12 = [ r² - 6] / 2 R.H.S is purely rational , whereas , L.H.S is irrational. This is a contradiction. This means that our assumption was wrong. Hence , √3 + √4 is irrational. #Lets assume that :#√3 + √4 is rational.#√3 + √4 = r , where r is rational#Squaring both sides , we get#[√3 + √4 ]² = r²#3 + 2√12 + 4 = r²#7 + 2√12 = r²#2√12 = r² - 6#√12 = [ r² - 6] / 2*R.H.S is purely rational , whereas , L.H.S is irrational.This is a contradiction.This means that our assumption was wrong.*Hence , √3 + √4 is irrational. *note.. # it is for separating steps.... Lets assume that :√3 + √4 is rational.√3 + √4 = r , where r is rationalSquaring both sides , we get =[√3 + √4 ]² = r² =3 + 2√12 + 4 = r² =7 + 2√12 = r² =2√12 = r² - 6 =√12 = [ r² - 6] / 2*R.H.S is purely rational , whereas , L.H.S is irrational.This is a contradiction.This means that our assumption was wrong.*Hence , √3 + √4 is irrational. Lets assume that :√3 + √4 is rational.√3 + √4 = r , where r is rationalSquaring both sides , we get[√3 + √4 ]² = r²3 + 2√12 + 4 = r²7 + 2√12 = r²2√12 = r² - 6√12 = [ r² - 6] / 2R.H.S is purely rational , whereas , L.H.S is irrational.This is a contradiction.This means that our assumption was wrong.Hence , √3 + √4 is irrational. |
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