Here's one example of where the difference between rational numbers and irrational numbers matters. Consider a circle of circumference $1$ (in any units you choose), and suppose we have an ant (of infinitesimal size, of course) on the circle that moves forward by $f$ instantaneously once per second. Then the ant will return to its starting point if and only if $f$ is a rational number.
Maybe that was a little contrived. How about this instead? Consider an infinite square lattice with a chosen point $O$. Choose another point $P$ and draw the line segment $O P$. Pick an angle $\theta$ and draw a line $L$ starting from $O$ so that the angle between $L$ and $O P$ is $\theta$. Then, the line $L$ passes through a lattice point other than $O$ if and only if $\tan \theta$ is rational.
In general the difference between rational and irrational becomes most apparent when you have some kind of periodicity in space or time, as in the examples above.
Yes, you could instead (uniquely!) write $\,p = 2^j a\,$ and $\,q = 2^k b\,$ for $\,a,b\,$ odd and then deduce a contradiction by comparing the number of factors of $2,\,$ viz. $\,p^2\,$ has an even number of $2$'s but $\,2q^2\,$ has an odd number of $2$'s, a contradiction. This method uses only the very easily-proved existence and uniqueness of $2$-factorizations, i.e. representations of the form $\,2^j n,\,$ with $\,n\,$ odd (versus the much more powerful, and much more difficult to prove Fundamental Theorem of Arithmetic = existence and uniqueness of arbitrary prime factorizations).
As for your second question, yes, every fraction $\,A/B\,$ can be written with coprime numerator and denominator $\,a/b\,$ simply by cancelling their gcd $\,c,\,$ i.e. $\,A/B = ca/cb = a/b.\,$ By the maximality ("greatest") property of the gcd it follows that $\,d =\gcd(a,b) = 1,\,$ else $\,cd\,$ would be a common divisor of $\,A,B\,$ larger than the greatest common divisor $\,c=\gcd(A,B),\,$ contradiction. However, as above, it suffices to cancel only common factors of $\,2,\,$ so no knowledge of gcds is required.
Best Answer
It's fine until you reach that equality $4^m=6^n$. But you can't just divide by their common factor $2$. Are you dividing by $2^m$ or by $2^n$?
You can say that, since $n>0$, then $3\mid6^n$. But $3\nmid4^m$.