Is
√2+√3/2
a rational number between √2 and √3?
Answers
Step-by-step explanation:
------------------------------
Given √2 + √3
To prove: √2 + √3 is an irrational number.
Proof:
Letus assume that √2 + √3 is a rational number.
So it can be written in the form a/b
√2 + √3 = a/b
Here a and b are coprime numbers and b ≠ 0
Solving
√2 + √3 = a/b
√2 = a/b – √3
On squaring both the sides we get,
=> (√2)2 = (a/b – √3)2
We know that
(a – b)2 = a2 + b2 – 2ab
So the equation (a/b – √3)2 can be written as
(a/b – √3)2 = a2/b2 + 3 – 2 (a/b)√3
Substitute in the equation we get
2 = a2/b2 + 3 – 2 × √3 (a/b)
Rearrange the equation we get
a2/b2 + 3 – 2 = 2 × √3 (a/b)
a2/b2 + 1 = 2 × √3 (a/b)
(a2 + b2)/b2 × b/2a = √3
(a2 + b2)/2ab = √3
Since, a, b are integers, (a2 + b2)/2ab is a rational number.
√3 is a rational number.
It contradicts to our assumption that √3 is irrational.
∴ our assumption is wrong
Thus √2 + √3 is irrational.
hope this helps you
Answer:
"
Given √2 + √3
To prove: √2 + √3 is an irrational number.
Proof:
Letus assume that √2 + √3 is a rational number.
So it can be written in the form a/b
√2 + √3 = a/b
Here a and b are coprime numbers and b
#0
Solving
√2 + √3 = a/b
√2 + √3 = a/b
√2 = a/b - √3
On squaring both the sides we get,
=> (√2)2 = (a/b - √3)2
We know that
(a - b)2 = a2 + b2 - 2ab
So the equation (a/b - √3)2 can be written as
(a/b - √3)2 = a2/b2 + 3-2 (a/b)√3
Substitute in the equation we get
2 = a2/b2 + 3 - 2 x √3 (a/b)
Rearrange the equation we get
a2/b2 + 3 - 2 = 2 × √3 (a/b)
a2/b2 + 1 = 2 × √3 (a/b)
(a2 + b2)/b2 × b/2a = √3
(a2 + b2)/2ab = √3
Since, a, b are integers, (a2 + b2)/2ab is a rational number.
√3 is a rational number.
It contradicts to our assumption that √3 is irrational.
.. our assumption is wrong
Thus √2 + √3 is irrational. "
Step-by-step explanation:
sry for deleting ur questions :(
but some one told me to do it