How to prove the irrationality of a number by contradiction method? Also, prove that sqrt2 and sqrt 7 are irrational.
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To prove : sqrt 2 is irrational
Assume sqrt 2 is rational.
So, sqrt can be written in the form of p/q, where p and q are integers, q is not 0, and p and q are co prime.
sqrt 2 = p/q
2 = p^2/q^2
2*q^2=p^2
Thus, p^2 is even, as 2*q^2 is even. This can only be true if p is even. So, p is even. We can write p=2s.
Now,
2*q^2=(2s)^2
2*q^2=4*s^2
q^2=(4*s^2)/2
q^2=2*s^2
Thus, q^2 is even, as 2*s^2 is even. This can only be true if q is even. So, q is even. We can write p=2r.
But now, p and q have a common factor, that is, 2.
This contradicts our assumption.
Hence, sqrt 2 cannot be rational.
Hence, sqrt 2 is irrational.
Assume sqrt 2 is rational.
So, sqrt can be written in the form of p/q, where p and q are integers, q is not 0, and p and q are co prime.
sqrt 2 = p/q
2 = p^2/q^2
2*q^2=p^2
Thus, p^2 is even, as 2*q^2 is even. This can only be true if p is even. So, p is even. We can write p=2s.
Now,
2*q^2=(2s)^2
2*q^2=4*s^2
q^2=(4*s^2)/2
q^2=2*s^2
Thus, q^2 is even, as 2*s^2 is even. This can only be true if q is even. So, q is even. We can write p=2r.
But now, p and q have a common factor, that is, 2.
This contradicts our assumption.
Hence, sqrt 2 cannot be rational.
Hence, sqrt 2 is irrational.
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Answer:
give 3 examples of euclidian geometry
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