prove that √2 is irrational tell me fast
Answers
Answered by
7
hope it helps u............
Attachments:
![](https://hi-static.z-dn.net/files/d18/f6a11f94f628889eb6d02f5dacca4233.jpg)
brainly11422:
thanks
Answered by
1
Hey there!
Prove that √2 is irrational
Let us assume the contrary, that is √2 is rational.
Then,
![\sqrt{2} = \frac{a}{b} \sqrt{2} = \frac{a}{b}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+%3D+%5Cfrac%7Ba%7D%7Bb%7D+)
where a and b are co-primes and b≠0
![2 = \frac{a {}^{2} }{b {}^{2} } 2 = \frac{a {}^{2} }{b {}^{2} }](https://tex.z-dn.net/?f=2+%3D+%5Cfrac%7Ba+%7B%7D%5E%7B2%7D+%7D%7Bb+%7B%7D%5E%7B2%7D+%7D+)
(by squaring both sides)
![a {}^{2} = 2b {}^{2} a {}^{2} = 2b {}^{2}](https://tex.z-dn.net/?f=a+%7B%7D%5E%7B2%7D+%3D+2b+%7B%7D%5E%7B2%7D+)
2 is a factor of a ²,
so, 2 is also a factor of a
![a = 2c a = 2c](https://tex.z-dn.net/?f=a+%3D+2c)
![(2c) {}^{2} = 2b {}^{2} (2c) {}^{2} = 2b {}^{2}](https://tex.z-dn.net/?f=%282c%29+%7B%7D%5E%7B2%7D+%3D+2b+%7B%7D%5E%7B2%7D+)
![4c {}^{2} = 2b {}^{2} 4c {}^{2} = 2b {}^{2}](https://tex.z-dn.net/?f=4c+%7B%7D%5E%7B2%7D+%3D+2b+%7B%7D%5E%7B2%7D+)
![2c {}^{2} = b {}^{2} 2c {}^{2} = b {}^{2}](https://tex.z-dn.net/?f=2c+%7B%7D%5E%7B2%7D+%3D+b+%7B%7D%5E%7B2%7D+)
So, b is a factor of 2
Since a and b are having 2 as common factor, this contradicts our assunption that a and b are co-primes.
So, √2 is not rational
Hence, √2 is irrational.
Hope it helps!
Prove that √2 is irrational
Let us assume the contrary, that is √2 is rational.
Then,
where a and b are co-primes and b≠0
(by squaring both sides)
2 is a factor of a ²,
so, 2 is also a factor of a
So, b is a factor of 2
Since a and b are having 2 as common factor, this contradicts our assunption that a and b are co-primes.
So, √2 is not rational
Hence, √2 is irrational.
Hope it helps!
Similar questions
Hindi,
7 months ago
Computer Science,
7 months ago
Hindi,
1 year ago
Sociology,
1 year ago
History,
1 year ago
Social Sciences,
1 year ago