Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 7D (Numbers and Sets)

If $q=\sqrt{2}+\sqrt{3}$ were a rational number, then $q^2=5+2\sqrt{6}$ would make $\sqrt{6}$ rational. Write $\sqrt{6}=a/b$ in lowest terms. Then $a^2=6b^2$, which is impossible by unique factorization: the exponent of the prime $2$ is even on the left and odd on the right. Therefore

The exponential function series gives Euler's number $e={\sum }_{k=0}^{\infty }1/k!$. Suppose $e=a/b$ with integers $a,b$ and $b>0$. Choose $n\ge \max (b,2)$. Since $b$ divides $n!$, the numberis an integer. But its positive tail satisfiesThe inequality is strict because the factors after the first exceed $n+1$ when $j\ge 2$. No integer lies strictly between zero and one. This proves the irrationality of e:

For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

If ${\log }_{2}3=a/b$ were a rational number, choose positive integers $a,b$ because ${\log }_{2}3>0$. Exponentiation gives $2^a=3^b$, but the left side is even and the right side is odd. ThusThis is the basic irrational logarithm criterion: a rational logarithm would force positive integer powers of its base and argument to coincide.