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Prove that the √5 is irrational

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Prove that the √5 is irrational

Posted by Chiluka Sriteja 1 year, 2 months ago

CBSE > Class 10 > Mathematics

3 answers

Priyanshu Singh 1 year, 2 months ago

assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.

Gunpreet Kaur 1 year, 2 months ago

lets assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.

Gunpreet Kaur 1 year, 2 months ago

lets assume that √5 is rational number √5=a/b (here 'a' and 'b' are co prime) b√5= a Square on both sides (b√5)^2 =(a)^2 5b^2 = a^2.............(i) If a^2 is divisible by 5 than it means that 5 is also divisible by a. Now, let there be another integer 'c' therefore, a=5c square on both side a^2 =25c^2 5b^2 =25c^2 {from equation (i)} then, b^2 =5 c^2 b^2/5 = c^2 If b^2 is divisible by 5 than it means b is also divisible by 5 . Therefore 'a' and 'b' has common factor 5 . But , we ensure that 'a' and 'b' are co prime. Therefore, our assumption is wrong that √5 is rational number. Contradiction occurs. Therefore √5 is irrational numbers.

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