proove that root 5is irrational
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Answered by
22
Hey!
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Let assume that √5 is a rational number
So we can find coprime integers a and b
b ≠ 0
√5 = a ÷ b
√5b = a
Squarring both sides
(√5b)² = a²
5b² = a²
5 divides a² and 5 also divides a (Therom)
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Now :-
a = 5c for some integer c
5b² = (5c)²
5b² = 25c²
b² = 5c²
_______________________________
5 divides b² than 5 also divides b (Therom)
A and b have a common factor 5 other than 1 which is contrary to the fact that a and b are Co Prime.
Our supposition is not correct
Hence √5 is an irrational number
_____________________________________________________________________________________________
Regards :)
Cybary
Be Brainly :)
_____________________________________________________________________________________________
Let assume that √5 is a rational number
So we can find coprime integers a and b
b ≠ 0
√5 = a ÷ b
√5b = a
Squarring both sides
(√5b)² = a²
5b² = a²
5 divides a² and 5 also divides a (Therom)
_______________________________
Now :-
a = 5c for some integer c
5b² = (5c)²
5b² = 25c²
b² = 5c²
_______________________________
5 divides b² than 5 also divides b (Therom)
A and b have a common factor 5 other than 1 which is contrary to the fact that a and b are Co Prime.
Our supposition is not correct
Hence √5 is an irrational number
_____________________________________________________________________________________________
Regards :)
Cybary
Be Brainly :)
saddle:
hi bro how do u do
Answered by
21
Heya !!!
If possible , suppose ✓5 is rational Number.
✓5 = A/B , where a and b are Integers and B not equal 0.
Let A and B have some common factor other than 1, then Divides A and B by that common factor .
Squaring both sides,
✓5 = A/B
(✓5)² = (A/B)²
5 = A²/B²
A² = 5B² -------(1)
5 Divides A² => 5 Divides A.
Let A = 2K , where K is some integer.
Putting A = 2K in equation (1)
25K² = 5B² => B² = 5K²
=> 5 Divides B² => 5 Divides B.
We see that A and B have common factor other than 5. Which is contradiction to the fact that a and b are Co prime.
Therefore,
Our SUPPOSITION is wrong that ✓5 is rational Number.
Hence,
✓5 is irrational Number.
HOPE IT WILL HELP YOU..... :-)
If possible , suppose ✓5 is rational Number.
✓5 = A/B , where a and b are Integers and B not equal 0.
Let A and B have some common factor other than 1, then Divides A and B by that common factor .
Squaring both sides,
✓5 = A/B
(✓5)² = (A/B)²
5 = A²/B²
A² = 5B² -------(1)
5 Divides A² => 5 Divides A.
Let A = 2K , where K is some integer.
Putting A = 2K in equation (1)
25K² = 5B² => B² = 5K²
=> 5 Divides B² => 5 Divides B.
We see that A and B have common factor other than 5. Which is contradiction to the fact that a and b are Co prime.
Therefore,
Our SUPPOSITION is wrong that ✓5 is rational Number.
Hence,
✓5 is irrational Number.
HOPE IT WILL HELP YOU..... :-)
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