Prove That √7 Is Irrational Number
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Let us assume that root7 is rational so
P/q where q is not equal to zero is a prime odd number
Root7/1 =p/q. Where p and q are coprime
Therefore p=qroot7
Now square in both side
P^2=(qroot5)^2
P^2=q^2*7
Therefore 7 divided p^2 and so 7 divided p (Theron 1.3)
Let p= 7c
(7c)^2=7q^2
49c^2=7q^2
49/7=q^2
7c^2=q^2. 7 divided q^2 so 7 divided q (1.3 Theron)
Therefore p and q have at least 7 as a common factor but this contradicts the fact that p and q have no common factor other than 1
This contradiction has arisen because of our wrong assumption that root 7 is rational so,we conclude that root 7 is irrational
Hope you understand
P/q where q is not equal to zero is a prime odd number
Root7/1 =p/q. Where p and q are coprime
Therefore p=qroot7
Now square in both side
P^2=(qroot5)^2
P^2=q^2*7
Therefore 7 divided p^2 and so 7 divided p (Theron 1.3)
Let p= 7c
(7c)^2=7q^2
49c^2=7q^2
49/7=q^2
7c^2=q^2. 7 divided q^2 so 7 divided q (1.3 Theron)
Therefore p and q have at least 7 as a common factor but this contradicts the fact that p and q have no common factor other than 1
This contradiction has arisen because of our wrong assumption that root 7 is rational so,we conclude that root 7 is irrational
Hope you understand
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