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By following the method of the proofs that and are irrational, would you expect to be able to prove that is also irrational? If you could, then find your mistake.
The pattern of the proof that square-root of 3 is not rational will be followed, but 3 will be replaced by 4.
We will try to prove by contradiction that p, the square-root of 4 (), is irrational (although we know that it is rational):
(1.1.1.4) |
Let's assume that p is a rational number and is represented in the reduced form:
(1.1.1.5) |
From (1.1.1.5) it follows that:
(1.1.1.6) |
If 4 is a factor of n it means that
(1.1.1.7) |
where k is another integer, then (1.1.1.6) can hold. On the other hand if 4 is not a factor of n, then there are three (3) possibilities:
(1.1.1.8) |
therefore n² does not satisfy (1.1.1.6). Really??? What about the second possibility of (1.1.1.8)? We found actually that if 2 is a factor of n (n is even), (1.1.1.6) holds but (1.1.1.7) does not have to.
Indeed we know that n=2 and m=1 corresponds to the square-root of 4 .
The goal of this exercise is to find the mistake when trying to deduce that the square-root of 4 is irrational. If you could not find that, your score is zero
Otherwise the score is 10.