Problems from Definition of irrational numbers

Development:

1. Let's suppose that $\sqrt{7}={\displaystyle \frac{p}{q}}$ 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 $\sqrt{7}doesnotexist.$

1. If $3\pi$ was rational we would have $3\pi ={\displaystyle \frac{p}{q}}$, where $p$ and $q$ are integers. Then we would have $\pi ={\displaystyle \frac{p}{3q}}$ 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.

Hide solution and development