prove that 1+1√2 is irrational with contradictory method
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prove that 1+1√2 is irrational with contradictory method
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Question :-
prove that 1 + 1√2 is irrational number with contradictory method .
Answer :-
Given :-
1 + 1√2
Required to prove :-
1 + 1√2 is an irrational number ?
Condition mentioned :-
By contradictory method
Conditions used :-
Here conditions refer to the properties ;
Similarly ,
If a divides q²
a divides q also .
At Last,
An irrational number is not equal to a rational number
number Q' ≠ Q
Here, Q' ( Q dash ) represents irrational numbers .
Here, Q' ( Q dash ) represents irrational numbers .Q represents rational numbers .
Proof :-
Given :-
1 + 1√2
Let's assume on the contradictory that 1 + 1√2 is a rational number .
So, equal the number with p/q .
( where p, q are integers , q ≠ 0 and p, q are co-primes )
[tex]\longrightarrow{\tt{ 1 + 1 \sqrt{2} = \dfrac{p}{q}}}[/tex]
Transpose +1 to the right side
[tex]\longrightarrow{\tt{1 \sqrt{2} = \dfrac{p}{q} - \dfrac{1}{1}}}[/tex]
By taking LCM on the right side
[tex]\longrightarrow{\tt{1\sqrt{2} = \dfrac{p - q}{q}}}[/tex]
So, here we can take 1√2 as √2
Since, any number divided or multiplied with 1 will give back the same number .
Hence,
[tex]\boxed{\tt{ \dfrac{p-q}{q} \; is \; a\; rational \; number}}[/tex]
[tex]\Rightarrow{\tt{ \sqrt{2} = \dfrac{p - q}{q}}}[/tex]
We know that
√2 is an irrational number .
But in the question it is not mentioned so we have to prove that √2 is an irrational number .
[tex]\rule{250}{6}[/tex]
So,
Lets assume on the contradictory that √2 is a rational number
Which is equal to a by b .
( where a, b are integers , b ≠ 0 and a, b are co-primes )
So,
[tex]\mathrm{ \sqrt{2} = \dfrac{a}{b}}[/tex]
By cross multiplication we get,
√2b = a
Squaring on both sides
( √2b )² = ( a )²
2b² = a²
Now recall the fundamental theorem of arithmetic
According to which ,
If a divides q²
Then, a divides q also .
So,
Here,
2b² = a²
2 divides a²
So, 2 divides a also .
However,
Let take the value of a as 2k
Substitute this in the above one .
So,
√2b = a
√2b = 2k
Squaring on both sides
( √2b)² = ( 2k )²
2b² = 4k²
[tex]\rm{ b^2 = \dfrac{4k^2}{2}}[/tex]
[tex]\implies{\rm{ b^2 = 2k^2 }}[/tex]
[tex]\implies{\rm{ 2k^2 = b^2 }}[/tex]
Here,
2 divides b²
So, 2 divides b also .
From the above we can conclude that ,
a, b have common factor 2 .
But according to the condition ,
a, b are co-primes which means a, b should have 1 as the common factor .
So,
This contradiction is due to the wrong assumption that √2 is a rational number .
So, our assumption is wrong .
Hence,
√2 is an irrational number
[tex]\rule{250}{6}[/tex]
From the above we came to know that √2 is an irrational number .
Hence,
[tex]\huge{\boxed{\sf{ \sqrt{2} \neq \dfrac{ p - q}{q}}}}[/tex]
This is because ,
An irrational number is not equal to a rational number
so,
This contradiction is due to the wrong assumption that 1 + 1√2 is a rational number .
Our assumption is wrong .
Hence
1 + 1√2 is an irrational
number .
Hence proved .