Applications of prime factorisation

Lemma 6.6.1   If $n \in \mathbb{N}$ is not a square number then $\sqrt{n}$ is irrational.

Proof. Suppose $\sqrt{n} = \frac{a}{b}$, with $(a,b) = 1$. Then $n b^2 = a^2$. If $b > 1$ then let $p$ be a prime dividing $b$. Thus $p \mid a^2$ and so $p \mid a$, which is impossible as $(a,b) = 1$. Thus $b = 1$ and $n = a^2$. $\qedsymbol$

This lemma can also be stated: ``if $n \in N$ with $\sqrt{n} \in \mathbb{Q}$ then $\sqrt{n} \in \mathbb{N}$''.

Definition 6.6.2   A real number is algebraic if it satisfies a polynomial equation with coefficients in $\mathbb{Z}$. Real numbers which are not algebraic are transcendental

For instance $\pi$ and $e$ are transcendental (the proof is fairly hard, especially for $\pi$). Most reals are transcendental; this can be proven using the notion of a countable set and we will do it later (v.i. Theorem thm transcendental numbers -> uncountable). If the rational $\frac{a}{b}$ ( with $(a,b) = 1$ ) satisfies a polynomial with coefficients in $\mathbb{Z}$ then

$$c_n a^n + c_{n-1} a^{n-1} b + \dots b^n c_0 = 0$$

so $b \mid c_n$ and $a \mid c_0$. In particular if $c_n = 1$ then $b = 1$, which is stated as ``algebraic integers which are rational are integers''. Note that if $a = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}$ and $b = p_1^{\beta_1} p_2^{\beta_2} \dots p_k^{\beta_k}$ with $\alpha_i, \beta_i \in \mathbb{N}_0$ then $(a,b) = p_1^{\gamma_1} p_2^{\gamma_2} \dots p_k^{\gamma_k}$ and $[a,b] = p_1^{\delta_1} p_2^{\delta_2} \dots p_k^{\delta_k}$, $\gamma_i = \min \{\alpha_i, \beta_i\}$ and $\delta_i = \max \{ \alpha_i, \beta_i \}$. Major open problems in the area of prime numbers are the Goldbach conjecture (``every even number greater than two is the sum of two primes'') and the twin primes conjecture (``there are infinitely many prime pairs $p$ and $p+2$'').