Problems from Definition of irrational numbers

Development:

1. Let's suppose that $$\sqrt{7}=\dfrac{p}{q}$$ where $$p$$ and $$q$$ are integers without factors in common. We multiply by $$q$$ and raise the expression to the square, obtaining; $$$7q^2=p^2$$$

If we do the factorization in prime numbers we see that on the left side there is a odd number of sevens and on the right side an even number. As such, we can say that a rational expression of $$\sqrt{7} does not exist.$$

1. If $$3\pi$$ was rational we would have $$3\pi=\dfrac{p}{q}$$, where $$p$$ and $$q$$ are integers. Then we would have $$\pi=\dfrac{p}{3q}$$ and $$\pi$$ would be rational, which is clearly false.

So, we can say that $$3\pi$$ is not rational.

Solution:

1. $$\sqrt{7}$$ is not rational.
2. $$3\pi$$ is not rational.

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