prove (3-root 7) square is an irratational number
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2
let us assume that 3-√7 is rational number
therefore it must be having coprime nos a and b
3-√7=a/b
3-a/b=√7
but√7 is irrational number
therefore 3-√7 is irrational number
therefore it must be having coprime nos a and b
3-√7=a/b
3-a/b=√7
but√7 is irrational number
therefore 3-√7 is irrational number
klaKarthik:
accutualy it is whole square
Answered by
17
Let √7 be a rational number, then
√7=p/q ,where p, q are integers, q is not equals to zero and p, q have no common factors (except 1).
or, 7 =p^2/q^2
or, p^2=7q^2. ...(i)
As 7 divides 7q^2, so 7 divides p^2 but 7 is prime.
Let p=7m, where m is an integer
Substituting the value of p in (i) we get
(7m)^2= 7q^2
or, 49m^2= 7q^2
or, 7m^2= q^2
As 7 divides 7m^2, so 7 divides q^2 but 7 is prime or, 7 divides q
Thus p and q have a common factor 7.
This contradicts that p and q have no common factors (except 1).
Hence, √7 is an irrational number.
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