Prove √5 is irrational
jaslam5566p4r0d6:
(5)^1/2 so 5 can't be split into its part like (4)^1/2= (2X2)^1/2=2 so hence proved its a irrational number
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✓5 is a irrational number let we assume a Nd b is a positive integer there fore
✓5=a/b
squaring bhot side
✓5=A2/b2
5=A2/b2
5b2=A2
5/A2
5/a
let c is a also some integers
b2=5/c2
b2=25/c2
5/b2
5/b
but their is a contradiction √5 is a rational number
✓5=a/b
squaring bhot side
✓5=A2/b2
5=A2/b2
5b2=A2
5/A2
5/a
let c is a also some integers
b2=5/c2
b2=25/c2
5/b2
5/b
but their is a contradiction √5 is a rational number
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HI
Proof:
Let us assume to the contrary that √5 is rational
Therefore, √5 = a/b , where a and b are coprime integers and b ≠ 0
Squaring on both sides,
5 = a²/b²
5b² = a² ......(1)
Since, a² is divisible by 5,
Therefore, a is divisible by 5
Let a = 5c and substitute in eq(1)
5b² = (5c)²
5b² = 25c²
b² = 5c²
Since, b² is divisible by 5,
Therefore, b is divisible by 5
➡️ a and b have at least one common factor i.e. 5
This contradicts the fact that a and b are coprime.
➡️ This contradiction has arisen due to our incorrect assumption that √5 is rational.
Therefore, we conclude that √5 is irrational.
Hence Proved !
Hope it proved to be beneficial....
Proof:
Let us assume to the contrary that √5 is rational
Therefore, √5 = a/b , where a and b are coprime integers and b ≠ 0
Squaring on both sides,
5 = a²/b²
5b² = a² ......(1)
Since, a² is divisible by 5,
Therefore, a is divisible by 5
Let a = 5c and substitute in eq(1)
5b² = (5c)²
5b² = 25c²
b² = 5c²
Since, b² is divisible by 5,
Therefore, b is divisible by 5
➡️ a and b have at least one common factor i.e. 5
This contradicts the fact that a and b are coprime.
➡️ This contradiction has arisen due to our incorrect assumption that √5 is rational.
Therefore, we conclude that √5 is irrational.
Hence Proved !
Hope it proved to be beneficial....
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