Show 4 - √3 is irrational
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Let us assume that 4-√3 is rational
4-√3=a/b(a and b are integers and b not equal to 0)
-√3 = a/b-4
√3 = -(a-4b/b)
√3 = -a+4b/b
Since a and b are integers and b not equal to 0 -a+4b/b is rational.√3,which is equal to -a+4b/b must be rational.But this contradicts the fact that √3 is irrational.Hence our assumption is wrong and (4-√3) is irrational.
Hence proved.
4-√3=a/b(a and b are integers and b not equal to 0)
-√3 = a/b-4
√3 = -(a-4b/b)
√3 = -a+4b/b
Since a and b are integers and b not equal to 0 -a+4b/b is rational.√3,which is equal to -a+4b/b must be rational.But this contradicts the fact that √3 is irrational.Hence our assumption is wrong and (4-√3) is irrational.
Hence proved.
Answered by
0
Hey there !!
To prove :-
4 - √3 is irrational
Lets assume that 4 - √3 is rational
Let ,
4 - √3 = r , where r is rational
4 - r = √3
Here ,
its clear that LHS is purely rational. But on the other hand , RHS is irrational.
This is a contradiction .
Hence , our assumption was wrong .
Therefore ,
4 - √3 is irrational
To prove :-
4 - √3 is irrational
Lets assume that 4 - √3 is rational
Let ,
4 - √3 = r , where r is rational
4 - r = √3
Here ,
its clear that LHS is purely rational. But on the other hand , RHS is irrational.
This is a contradiction .
Hence , our assumption was wrong .
Therefore ,
4 - √3 is irrational
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