Prove that √2 is irrational using the fundamental theorem of arithmetic.
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let √2 be a rational no. then it can be expressed in the form of p/q where p and q are co primes and q is not equal to zero, then:-
√2= p/q
squaring both sides
2 = (p/q)^2
2q^2 =p^2
q^2= p^2/2...eq 1
2 is the factor of p^2 and hence of p also.
taking p = 2c then
q^2 = (2c)^2/2
q^2 = 4c^2/2
q^2= 2c^2
q^2/2 = 2c^2....eq 2
2 is a factor of q^2 and q also
therefore 2 is a factor of both p and q
but p and q are co primes
here comes a contradiction
hence √2 is irrational.
√2= p/q
squaring both sides
2 = (p/q)^2
2q^2 =p^2
q^2= p^2/2...eq 1
2 is the factor of p^2 and hence of p also.
taking p = 2c then
q^2 = (2c)^2/2
q^2 = 4c^2/2
q^2= 2c^2
q^2/2 = 2c^2....eq 2
2 is a factor of q^2 and q also
therefore 2 is a factor of both p and q
but p and q are co primes
here comes a contradiction
hence √2 is irrational.
Deepayu:
2 is a factor of p2 because it gives us an answer i.e. q2
and 2 is a factor of p because it is a factor of p2
for eg 2 is a factor of 100 hence 2 is a factor of 10 as well
i hope you have cleared your doubt...
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