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Prove that p, the square-root of 3 (), is irrational. ( hint: see the example for square-root of 2 )
The pattern of the proof that square-root of 2 is not rational will be followed, but 2 will be replaced by 3.
We will prove by contradiction that p, the square-root of 3 (), is irrational:
(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 3 is a factor of n it means that
(1.1.1.7) |
where k is another integer, then (1.1.1.6) is true. On the other hand if 3 is not a factor of n, then there are two possibilities:
(1.1.1.8) |
therefore n² does not satisfy (1.1.1.6). From (1.1.1.6) and (1.1.1.7) one obtains:
(1.1.1.9) |
and therefore 3 is a factor of m, which is inconsistent with the assumption that p is a rational number represented in the reduced form.
The proof follows exactly the pattern for square-root of 2, except for the two possibilities in equation (1.1.1.8) instead of one before.
If your proof follows this one, you scored 10 points.
If your proof follows this one, but you missed the two possibilities of (1.1.1.8), your score is 3.
Otherwise your score is zero