Prove that root 7 is irrational
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To prove 2/root 7 is irrational, you have to show that root 7 is irrational, as you know that when a rational number is divided by irrational no. then the no. obtained is irrational. We can do this by method of contradiction. =>root 7 =p/q, where p and q are coprimes and q is not equal to 0.
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To prove 2/root 7 is irrational, you have to show that root 7 is irrational, as you know that when a rational number is divided by irrational no. then the no. obtained is irrational.
To prove- Root 7 is irrational.
Proof-
We can do this by method of contradiction.
Let root 7 is rational.
=> root 7 =p/q, where p and q are coprimes and q is not equal to 0.
=> 7=(p^2)/(q^2)
=>p^2= 7q^2
=>7 is factor of p^2
=> 7 is factor of p
=>p=7k, where k is a constant
=>p^2=49k^2
=>7q^2=49k^2
=>q^2=7k^2
=> 7 is also a factor of q^2 and thus a factor of q.
p and q have 7 as a common factor except 1.
This is a pure contradiction to the fact that p and q are coprimes i.e they have only 1 as the common factor.
So our assumption that root 7 was rational is wrong.
Root 7 is irrational.
So 2, a rational no. When divided by root 7 the answer is irrational.
To prove- Root 7 is irrational.
Proof-
We can do this by method of contradiction.
Let root 7 is rational.
=> root 7 =p/q, where p and q are coprimes and q is not equal to 0.
=> 7=(p^2)/(q^2)
=>p^2= 7q^2
=>7 is factor of p^2
=> 7 is factor of p
=>p=7k, where k is a constant
=>p^2=49k^2
=>7q^2=49k^2
=>q^2=7k^2
=> 7 is also a factor of q^2 and thus a factor of q.
p and q have 7 as a common factor except 1.
This is a pure contradiction to the fact that p and q are coprimes i.e they have only 1 as the common factor.
So our assumption that root 7 was rational is wrong.
Root 7 is irrational.
So 2, a rational no. When divided by root 7 the answer is irrational.
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