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Divisibility
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OBJECTIVES
At the end of this topic, as a pre-service teacher, you are
expected to:
A. discuss the definition and properties on divisibility;
B. define theorems related to divisibility; and
C. Solve problems of divisibility.
3
Definition. If a and b are integers with 𝒂 ≠ 𝟎,
we say that a divides b if there is an integer
c such that 𝒃 = 𝒂𝒄. If a divides b, we also
say that a is a divisor or factor of b and that
b is a multiple of a.
If a divides b, we write 𝒂|𝒃, while if a does NOT
divide b, we write 𝒂 ∤ 𝒃.
EXAMPLES
1. 𝟏𝟑 | 𝟏𝟖𝟐 (13 divides 182 or 13 is a divisor of 182)
2. −𝟐 | 𝟐𝟓𝟎
3. 𝟒 | 𝟏𝟎𝟎
4. 𝟔 ∤ 𝟒𝟒 (6 does not divide 44 or 6 is not a divisor of 44)
5. The divisors of 𝟏𝟐 are
± 𝟏, ±𝟐, ±𝟑, ±𝟒, ±𝟔, 𝒂𝒏𝒅 ± 𝟏𝟐
6. The divisors of 100 are
± 𝟏, ±𝟐, ±𝟒, , ±𝟓, ±𝟏𝟎, ±𝟐𝟎, ±𝟐𝟓, ±𝟓𝟎, 𝒂𝒏𝒅 ±
𝟏𝟎𝟎.
5
Theorem. If a, b, and c are integers with 𝒂|𝒃 and 𝒃|𝒄, then 𝒂|𝒄.
Proof. Because 𝒂|𝒃 and 𝒃|𝒄, there are integers e and f such that 𝒂𝒆 = 𝒃 and
𝒃𝒇 = 𝒄. Hence, 𝒄 = 𝒃𝒇 = 𝒂𝒆 𝒇 = 𝒂(𝒆𝒇), and we conclude that 𝒂|𝒄.
Example. Because 𝟏𝟏|𝟔𝟔 and 𝟔𝟔|𝟏𝟗𝟖, the previous theorem tells us that
𝟏𝟏|𝟏𝟗𝟖.
Theorem. If a, b, m, and n are integers, and if 𝒄|𝒂 and 𝒄|𝒃, then 𝒄|(𝒎𝒂 + 𝒏𝒃).
Proof. Because 𝒄|𝒂 and 𝒄|𝒃, there are integers e and f such that 𝒂 = 𝒄𝒆 and 𝒃 =
𝒄𝒇. Hence, 𝒎𝒂 + 𝒏𝒃 = 𝒎𝒄𝒆 + 𝒏𝒄𝒇 = 𝒄(𝒎𝒆 + 𝒏𝒇), Consequently, we see
that 𝒄|(𝒎𝒂 + 𝒏𝒃).
Example. As 𝟑|𝟐𝟏 and 𝟑|𝟑𝟑, previous theorem tells us that 3 divides 𝟓 ∗ 𝟐𝟏– 𝟑 ∗
𝟑𝟑 = 𝟏𝟎𝟓 − 𝟗𝟗 = 𝟔
6
The Division Algorithm
> Theorem. If a and b are integers such that 𝒃 >
𝟎, then there are unique integers q and r such
that 𝒂 = 𝒃𝒒 + 𝒓 with 𝟎 ≤ 𝒓 < 𝒃.
> In this theorem, q is the quotient and r the
remainder, a the dividend and b the divisor. We
note that a is divisible by b if and only if the
remainder in the division algorithm is 0
7
EXAMPLES
1. If 𝒂 = 𝟏𝟑𝟑 and 𝒃 = 𝟐𝟏, then 𝒒 = 𝟔 and 𝒓 = 𝟕, because
𝟏𝟑𝟑 = 𝟐𝟏 ∗ 𝟔 + 𝟕. We rewrite this as:
> 133
21
= 6 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
7
21
2. If 𝒂 = 𝟏𝟓𝟓 and 𝒃 = 𝟏𝟒, then 𝒒 = 𝟏𝟏 and 𝒓 = 𝟏,
because 𝟏𝟓𝟓 = 𝟏𝟒 ∗ 𝟏𝟏 + 𝟏. We rewrite this as:
> 155
14
= 11 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
1
14
8
EXAMPLES
3. If 𝒂 = 𝟏𝟒𝟕 and 𝒃 = 𝟏𝟔, then 𝒒 = 𝟗 and 𝒓 = 𝟑, because
𝟏𝟒𝟕 = 𝟏𝟔 ∗ 𝟗 + 𝟑. We rewrite this as:
> 147
16
= 9 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
3
21
9
The following examples display the quotient and remainder of a division.
1. Let 𝑎 = 1028 and 𝑏 = 34. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where
𝑞 =
1028
34
= 30 and
𝑟 = 1028 −
1028
34
∗ 34 = 1028 − 30 ∗ 34 = 8
Check: 1028 = 34 30 + 8
2. Let 𝑎 = −380 and 𝑏 = 75. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where
𝑞 =
−380
75
= −6 and
𝑟 = −380 −
−380
75
∗ 75 = −380 − −6 75 = 70.
Check: −380 = 75 −6 + 70
10
The following examples display the quotient and remainder of a division.
3. Let 𝑎 = −3562 and 𝑏 = −232. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b,
where 𝑞 =
−3562
−232
= 15 and
𝑟 = −3562 −
−3562
−232
∗ −232 = −3562 − 15 −232 = −82.
Check: −3562 = −232 15 − 82
4. Let 𝑎 = 36724 and 𝑏 = −3647. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b,
where 𝑞 =
36724
−3647
= −11 and
𝑟 = 36724 − −11 −3647 = −3393.
Check: 36724 = −3647 −11 − 3393
11
“The real fundamental
building blocks of the
universe and matter are
not indivisible particles,
but infinitely divisible
moments in time.”
― Khalid Masood
12
13
Thanks!
End of Presentation
Contact me at jobert.suarez16@gmail.com

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M108 - Number Theory - Divisibility

  • 2. DON’T FORGET! Be Punctual Be Dressed/Groomed Find a Distraction-free place Be Respectful and Positive Stay on Mute – Raise Hand if you want to Speak Keep your Video ON Stay Focused and on task
  • 3. OBJECTIVES At the end of this topic, as a pre-service teacher, you are expected to: A. discuss the definition and properties on divisibility; B. define theorems related to divisibility; and C. Solve problems of divisibility. 3
  • 4. Definition. If a and b are integers with 𝒂 ≠ 𝟎, we say that a divides b if there is an integer c such that 𝒃 = 𝒂𝒄. If a divides b, we also say that a is a divisor or factor of b and that b is a multiple of a. If a divides b, we write 𝒂|𝒃, while if a does NOT divide b, we write 𝒂 ∤ 𝒃.
  • 5. EXAMPLES 1. 𝟏𝟑 | 𝟏𝟖𝟐 (13 divides 182 or 13 is a divisor of 182) 2. −𝟐 | 𝟐𝟓𝟎 3. 𝟒 | 𝟏𝟎𝟎 4. 𝟔 ∤ 𝟒𝟒 (6 does not divide 44 or 6 is not a divisor of 44) 5. The divisors of 𝟏𝟐 are ± 𝟏, ±𝟐, ±𝟑, ±𝟒, ±𝟔, 𝒂𝒏𝒅 ± 𝟏𝟐 6. The divisors of 100 are ± 𝟏, ±𝟐, ±𝟒, , ±𝟓, ±𝟏𝟎, ±𝟐𝟎, ±𝟐𝟓, ±𝟓𝟎, 𝒂𝒏𝒅 ± 𝟏𝟎𝟎. 5
  • 6. Theorem. If a, b, and c are integers with 𝒂|𝒃 and 𝒃|𝒄, then 𝒂|𝒄. Proof. Because 𝒂|𝒃 and 𝒃|𝒄, there are integers e and f such that 𝒂𝒆 = 𝒃 and 𝒃𝒇 = 𝒄. Hence, 𝒄 = 𝒃𝒇 = 𝒂𝒆 𝒇 = 𝒂(𝒆𝒇), and we conclude that 𝒂|𝒄. Example. Because 𝟏𝟏|𝟔𝟔 and 𝟔𝟔|𝟏𝟗𝟖, the previous theorem tells us that 𝟏𝟏|𝟏𝟗𝟖. Theorem. If a, b, m, and n are integers, and if 𝒄|𝒂 and 𝒄|𝒃, then 𝒄|(𝒎𝒂 + 𝒏𝒃). Proof. Because 𝒄|𝒂 and 𝒄|𝒃, there are integers e and f such that 𝒂 = 𝒄𝒆 and 𝒃 = 𝒄𝒇. Hence, 𝒎𝒂 + 𝒏𝒃 = 𝒎𝒄𝒆 + 𝒏𝒄𝒇 = 𝒄(𝒎𝒆 + 𝒏𝒇), Consequently, we see that 𝒄|(𝒎𝒂 + 𝒏𝒃). Example. As 𝟑|𝟐𝟏 and 𝟑|𝟑𝟑, previous theorem tells us that 3 divides 𝟓 ∗ 𝟐𝟏– 𝟑 ∗ 𝟑𝟑 = 𝟏𝟎𝟓 − 𝟗𝟗 = 𝟔 6
  • 7. The Division Algorithm > Theorem. If a and b are integers such that 𝒃 > 𝟎, then there are unique integers q and r such that 𝒂 = 𝒃𝒒 + 𝒓 with 𝟎 ≤ 𝒓 < 𝒃. > In this theorem, q is the quotient and r the remainder, a the dividend and b the divisor. We note that a is divisible by b if and only if the remainder in the division algorithm is 0 7
  • 8. EXAMPLES 1. If 𝒂 = 𝟏𝟑𝟑 and 𝒃 = 𝟐𝟏, then 𝒒 = 𝟔 and 𝒓 = 𝟕, because 𝟏𝟑𝟑 = 𝟐𝟏 ∗ 𝟔 + 𝟕. We rewrite this as: > 133 21 = 6 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 7 21 2. If 𝒂 = 𝟏𝟓𝟓 and 𝒃 = 𝟏𝟒, then 𝒒 = 𝟏𝟏 and 𝒓 = 𝟏, because 𝟏𝟓𝟓 = 𝟏𝟒 ∗ 𝟏𝟏 + 𝟏. We rewrite this as: > 155 14 = 11 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 1 14 8
  • 9. EXAMPLES 3. If 𝒂 = 𝟏𝟒𝟕 and 𝒃 = 𝟏𝟔, then 𝒒 = 𝟗 and 𝒓 = 𝟑, because 𝟏𝟒𝟕 = 𝟏𝟔 ∗ 𝟗 + 𝟑. We rewrite this as: > 147 16 = 9 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 3 21 9
  • 10. The following examples display the quotient and remainder of a division. 1. Let 𝑎 = 1028 and 𝑏 = 34. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where 𝑞 = 1028 34 = 30 and 𝑟 = 1028 − 1028 34 ∗ 34 = 1028 − 30 ∗ 34 = 8 Check: 1028 = 34 30 + 8 2. Let 𝑎 = −380 and 𝑏 = 75. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where 𝑞 = −380 75 = −6 and 𝑟 = −380 − −380 75 ∗ 75 = −380 − −6 75 = 70. Check: −380 = 75 −6 + 70 10
  • 11. The following examples display the quotient and remainder of a division. 3. Let 𝑎 = −3562 and 𝑏 = −232. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where 𝑞 = −3562 −232 = 15 and 𝑟 = −3562 − −3562 −232 ∗ −232 = −3562 − 15 −232 = −82. Check: −3562 = −232 15 − 82 4. Let 𝑎 = 36724 and 𝑏 = −3647. Then 𝑎 = 𝑏𝑞 + 𝑟 with 0 ≤ 𝑟 <b, where 𝑞 = 36724 −3647 = −11 and 𝑟 = 36724 − −11 −3647 = −3393. Check: 36724 = −3647 −11 − 3393 11
  • 12. “The real fundamental building blocks of the universe and matter are not indivisible particles, but infinitely divisible moments in time.” ― Khalid Masood 12
  • 13. 13 Thanks! End of Presentation Contact me at jobert.suarez16@gmail.com