Elementary Number Theory - 1.3 Definitions and Propositions

Theorem: (Euclid)
There are infinitely many primes.


Proof:
This is a classic proof by contradiction. Assume that there are only finitely many primes, say \(n\) of them and denote them by \(\mathbb{P} = \{p_1,p_2,\ldots,p_n\}\). Consider the number \(m = p_1p_2 \ldots p_n +1\). If \(m\) is a prime, we have \(m \neq p_l\) for any \(l \in \{1,2,\ldots,n\}\). Hence, we have constructed a new prime not in \(\mathbb{P}\). This means either \(m\) is not a prime (or) there are infinitely many primes. Hence, if \(m\) is a prime, we have proved that there are infinitely many primes. If \(m\) is not a prime, then since \(m\) is also not an irreducible, and since we have assumed that only finitely many primes exist, we have that \(p_i | m\) for some \(p_i\). But we have \(m - p_1p_2 \ldots p_n = 1\). This means the \(p_i | 1\). CONTRADICTION. Hence, there are infinitely many primes.

Definition:
The greatest integer part of a real number \(\alpha\), denoted by \(\lfloor \alpha \rfloor\), is the greatest integer less than or equal to \(\alpha\) i.e. the unique integer \(m \in \mathbb{Z}\) such that \(m \leq \alpha \lt m+1\).


Proposition:
Suppose \(\alpha,\beta \in \mathbb{R}\).

1. We have \(\alpha-1 \lt \lfloor \alpha \rfloor \leq \alpha\) and \(0 \leq \alpha - \lfloor \alpha \rfloor \lt 1\).
• By definition, we have that \(\lfloor \alpha \rfloor \in \mathbb{Z}\) such that \(\lfloor \alpha \rfloor \leq \alpha \lt \lfloor \alpha \rfloor + 1\). Rearranging the inequalities, we get \(\alpha - 1 \lt \lfloor \alpha \rfloor \leq \alpha\). Again by rearranging, we get \(0 \leq \alpha - \lfloor \alpha \rfloor \lt 1\).
2. If \(\alpha \geq 0\), then \(\lfloor \alpha \rfloor\) counts the number of positive integers not exceeding \(\alpha\). In other words, \[\lfloor \alpha \rfloor = \sum_{1 \leq n \leq \alpha} 1.\]
• \(\sum_{1 \leq n \leq \alpha}\) 1 = \(\sum_{1 \leq n \leq \lfloor \alpha \rfloor} 1 + \sum_{\lfloor \alpha \rfloor \lt n \leq \alpha} 1\). Since we have \(0 \leq \alpha - \lfloor \alpha \rfloor \lt 1\), the contribution of the second term is zero as there is no integer \(n\) such that \(\lfloor \alpha \rfloor \lt n \leq \alpha\). The first term is nothing but \(\lfloor \alpha \rfloor\).
3. For every \(n \in \mathbb{Z}\), we have \(\lfloor \alpha + n \rfloor = \lfloor \alpha \rfloor + n\).
• Let \(\lfloor \alpha \rfloor = m \in \mathbb{Z}\). We have that \(m \leq \alpha \lt m+1\). Hence, we have \(m + n \leq \alpha + n \lt (m+n) + 1\). Hence, we get \(\lfloor \alpha +n \rfloor = m + n = \lfloor \alpha \rfloor + n\).
4. We have \(\lfloor \alpha \rfloor + \lfloor \beta \rfloor \leq \lfloor \alpha + \beta \rfloor \leq \lfloor \alpha \rfloor + \lfloor \beta + 1\).
• Let \(\lfloor \alpha \rfloor = m\) and \(\lfloor \beta \rfloor = n\). This means \(m \leq \alpha \lt m + 1\) and \(n \leq \beta \lt n + 1\). Adding the two, we get \(m + n \leq \alpha + \beta + m + n + 2\). Hence, we get \(\lfloor \alpha \rfloor + \lfloor \beta \rfloor \leq \lfloor \alpha + \beta \rfloor \leq \lfloor \alpha \rfloor + \lfloor \beta + 1\).
5. If \(\alpha \in \mathbb{Z}\), then \(\lfloor \alpha \rfloor + \lfloor -\alpha \rfloor = 0\). If \(\alpha \not \in \mathbb{Z}\), then \(\lfloor \alpha \rfloor + \lfloor -\alpha \rfloor = -1\).
• If \(\alpha \in \mathbb{Z}\), then \(-\alpha \in \mathbb{Z}\). Hence, we have \(\lfloor \alpha \rfloor = \alpha\) and \(\lfloor -\alpha \rfloor = - \alpha\). Hence, \(\lfloor \alpha \rfloor + \lfloor -\alpha \rfloor = \alpha + (- \alpha) = 0\). If \(\alpha \notin \mathbb{Z}\), then let \(m = \lfloor \alpha \rfloor\). Since \(\alpha \notin \mathbb{Z}\), we get \(m \lt \alpha \lt \alpha + 1\). Multiplying by negative one, we get \(-m-1 \lt -\alpha \lt -m \). This means \(\lfloor \alpha \rfloor = - m - 1\). Hence, we get \(\lfloor \alpha \rfloor + \lfloor -\alpha \rfloor = m + (-m-1) = -1\).
6. The number \(- \lfloor -\alpha \rfloor\) is the smallest integer not less than \(\alpha\).
• Let \(m = \lfloor \alpha \rfloor\). If \(\alpha \in \mathbb{Z}\), then \(m = \alpha\). We get \(-\alpha = -m \in \mathbb{Z}\). Hence, \(- \lfloor -\alpha \rfloor = m\) which is nothing but the smallest integer not less than \(\alpha\). If \(\alpha \notin \mathbb{Z}\), then \(m \lt \alpha \lt m + 1\). Hence, \(-m - 1 \lt -\alpha \lt -m \). Hence, we get \(\lfloor -\alpha \rfloor = - m - 1\). Hence, we get \(- \lfloor -\alpha \rfloor = m + 1\) which is nothing but the smallest integer not less than \(\alpha\).
7. If \(n \in \mathbb{Z}^+\), then \(\lfloor \lfloor \alpha \rfloor / n \rfloor = \lfloor \alpha/n \rfloor\).
• If \(\alpha \in \mathbb{Z}\), then it follows trivially since \(\lfloor \alpha \rfloor = \alpha\). If \(\alpha \notin \mathbb{Z}\), let \(m = \lfloor \alpha \rfloor\). We then have \(m = qn + r\) where \(r \in \{0,1,2,\ldots,n-1\}\). Hence, \(\alpha = qn + r + \{\alpha\}\) where \(\{\alpha\} \in [0,1)\). Note that \(r + \{\alpha\} \in (0,n)\). We then get \(\lfloor \lfloor \alpha \rfloor / n \rfloor = \lfloor \frac{qn+r}{n} \rfloor = q\). We also get \(\lfloor \alpha/n \rfloor = \lfloor (qn + r + \{\alpha\})/n \rfloor\) = q. Hence, we get the desired result.
8. The number \(\lfloor \alpha + \frac12 \rfloor\) is one of the two nearest integers to \(\alpha\). Furthermore, if these two integers differ from \(\alpha\) by the same value, then \(\lfloor \alpha + \frac12 \rfloor\) is the larger of these two integers.
• Follows immediately from the definition.
9. If \(\alpha \gt 0\) and \(n \in \mathbb{N}\), then \(\lfloor \frac{\alpha}{n} \rfloor\) is the number of positive integers not exceeding \(\alpha\) and which are multiples of \(n\).
• Follows from the second part of this problem.

Theorem:

Suppose that \(n \in \mathbb{N}\) and \(p\) is a prime. Then the largest integer \(k\) such that \(p^k | n!\) is given by \[k = \sum_{j=1}^{\infty} \left \lfloor \frac{n}{p^j} \right \rfloor .\]
Proof:
Let \(r_m\) be the highest power of \(p\) dividing \(m\). Then we get \(k = \displaystyle \sum_{m=1}^{n} r_m\). We have \(r_m = \displaystyle \sum_{l=1}^{\infty} 1_{\{p^l | m\}} \). Hence, we get \(k = \displaystyle \sum_{m=1}^{n} \sum_{l=1}^{\infty} 1_{\{p^l | m\}} = \sum_{l=1}^{\infty} \sum_{m=1}^{n} 1_{\{p^l | m\}} = \sum_{l=1}^{\infty} \left \lfloor \frac{n}{p^l} \right \rfloor\). Hence, the largest integer \(k\) such that \(p^k | n!\) is given by \[k = \sum_{j=1}^{\infty} \left \lfloor \frac{n}{p^j} \right \rfloor .\]