Answer:
the modified proof?
To modify Pythagoras' proof of the irrationality of the square root of 2 to prove the irrationality of the square root of 9 we would start by assuming that the square root of 9 is rational. This means we can write it as a fraction in the form \( \frac{a}{b} \ where \( a \) and \( b \) are integers with no common factors other than 1.
We can then follow the same steps as in Pythagoras' proof. First we square both sides to get \( 9 = \frac{a^2}{b^2} \). Multiplying both sides by \( b^2 \) gives us \( 9b^2 = a^2 \).
Here is where the modified proof breaks down. In Pythagoras' original proof he showed that the assumption of \( \frac{a}{b} \) being in its simplest form (with no common factors other than 1) led to a contradiction. However in the case of \( \sqrt{9} \ we can rewrite \( 9b^2 = a^2 \) as \( (3b)^2 = a^2 \).
This means that \( a \) is a multiple of 3 and we can rewrite it as \( a = 3c \) for some integer \( c \). Substituting this back into the equation gives us \( (3b)^2 = (3c)^2 \). Simplifying further we get \( 9b^2 = 9c^2 \).
Notice that this equation does not lead to a contradiction as in the original proof. In fact it implies that \( b = c \ which means that the fraction \( \frac{a}{b} \) is not necessarily in its simplest form. Therefore we cannot conclude that the square root of 9 is irrational using the modified proof based on Pythagoras' original proof for the square root of 2.
In summary the modified proof breaks down because the assumption of \( \frac{a}{b} \) being in its simplest form does not necessarily hold in the case of the square root of 9.