prove that √11 is irrational
Answers
Step-by-step explanation:
Firstly we assume that √11 is a rational number.
A rational number can be written in the form of p/q, where q ≠ 0 and p , q are positive number.
√11 = p/q ….( Where p & q are co prime number )
Square both sides, we get
11 = p²/q²
11 q² = p² ….(a)
p² is divisible by 11. So, p will also divisible by 11
Let p = 11 r ( Where r is any positive integer )
Squaring both sides
p² = 121r²
Putting this value in eqn(a)
11 q² = 121 r²
q² = 11 r²
q² is divisible by 11. So, q will also divisible by 11.
Since, p and q both are divisible by same number 11.
So, they are not co-prime .
Hence Our assumption is Wrong.
Therefore, √11 is an irrational number . Ans .
Step-by-step explanation:
To Prove : \tt{\sqrt{11}}
11
is irrational.
Proof :
Assume that \tt{\sqrt{11}}
11
is rational.
So,
\tt{\sqrt{11}= \dfrac{a}{b}} \: \: \: \gray{ \rm{(a \: and \: b \:are \: co - prime \: numbers)}}
11
=
b
a
(aandbareco−primenumbers)
\tt{\longrightarrow} \: \sqrt{11}b = a⟶
11
b=a
Squaring both the sides
\tt{\longrightarrow} \: {11b}^{2} = {a}^{2} - - - (1)⟶11b
2
=a
2
−−−(1)
As 11b² = a², so a = 11c. 11 is a factor of a.
Substitute the value of a in equation 1.
\tt{\longrightarrow} \: {11b}^{2} = {(11c)}^{2}⟶11b
2
=(11c)
2
\tt{\longrightarrow} \: {11b}^{2} = {121c^{2}}⟶11b
2
=121c
2
Divide both the sides by 11.
\tt{\longrightarrow} \: {b}^{2} = {11c^{2}}⟶b
2
=11c
2
11 is a factor of b.
Both a and b have 11 as their common factor.
This is a contradiction to the assumption of considering a and b co-prime numbers.
The contradiction arisen due to wrong assumption.
Therefore, \tt{\sqrt{11}}
11
is irrational.