MATH102-lecture-20220201

Divisibility cont. #

image_2022-02-01-08-57-48

There are many ways to represent 24, but there is only 1 way to represent it as a product of primes.

Theorem. Whenever we add a number to itself, the sum will be even.
Proof. Let first number be \( k \) , let the second number be \( k \) . So if \[\begin{aligned} k + k = 2k \end{aligned}\] and \( 2 | 2k \)

\( \square \)

Example
Prove that if \( 10 | a \) and \( 10 | b \) , then \( 10 | ma \pm nb \) , where \( a, b, m, n \in \mathbb{Z} \) .
Proof. \( 10 | a \) so \( a = 10k_1 \) , \( 10 | b \) so \( b = 10k_2 \) , where \( k_1, k_2 \) are unknown \( \in \mathbb{Z} \) . \[\begin{aligned} ma \pm nb &= m (10k_1) \pm n (10 k_2) \\ &= 10(m k_1 \pm n k_2) \end{aligned}\]

Note that \( (m k_1 \pm n k_2) \in \mathbb{Z} \) .

\( \square \)

Example
\( 7 \not | a \) and \( 7 \not | b \) . \( a + b \) may or may not be divisible by 7.

For example: \( 7 \not | 13 \) , \( 7 \not | 15 \) , but \( 7 | (13 + 15) \) .

For example: \( 7 \not | 15 \) , \( 7 \not | 16 \) , and \( 7 \not | (15 + 16) \) .

Example
If \( 7 | a \) and \( 7 \not | b \) , then it is certainly the case that \( 7 \not | a + b \) .
Proof. \( 7 | a \) so \( \frac{a}{7} \) is a whole number. \( 7 \not | b \) so \( \frac{b}{7} \) is not a whole number.

So \( \frac{a}{7} + \frac{b}{7} \) is not whole. Therefore, \( \frac{a + b}{7} \) is not a whole number.

So, by statement 5: \( 7 \not | a + b \) .

\( \square \)

Nmemonics for divisibility #

So, based on the examples above:

  1. IS + IS = IS
  2. IS + IS NOT = IS NOT
  3. IS NOT + IS NOT = IS or IS NOT
Example
\[\begin{aligned} 7^3 \cdot 19^{24} \not = 5^2 \cdot 11^{30} \end{aligned}\]

By the fundamental theory of arithmetic, each number only has 1 pf, therefore these number cannot equal each other.

Can \( 2^6 | 2^5\) ? No, because \( \frac{2^5}{2^6} = \frac{1}{2} \) , which is not a whole number.

Example
Prove that \( 5^3 \) cannot divide \( 3^2 \cdot 5^2 \cdot 11^3 \) .
Proof. Numbers that are divisible by \( 5^3 \) are of the form \( 5^3k \) .

There are two cases:

  1. \( k \) contains a power of 5, therefore the power of \( 5^3k \) will be greater than 3.
  2. \( k \) does not contain a power of 5, therefore the power of \( 5^3k \) is 3.

Both of these scenarios have too many 5s to go into numbers with a \( 5^2 \) in their prime factorization.

\( \square \)

Amount of divisors #

If \( N = 2^{11} \cdot 3 \cdot 19^{20} \) . We can multiply the exponents (+ 1) together to get the total amount of divisors: \( (11 + 1) (1 + 1) (20 + 1) = 25 \) .

image_2022-02-01-10-04-28 image_2022-02-01-10-11-52

  • the multiplication principle

image_2022-02-01-10-14-01