- The set of all even numbers
- The set of all books written about travel to Chile

- {1, 3, 9, 12}
- {red, orange, yellow, green, blue, indigo, purple}

*A* = the set of all even numbers
*B* = {2, 4, 6}
*C* = {2, 3, 4, 6}

*A* = {red, green, blue}
*B* = {red, yellow, orange}
*C* = {red, orange, yellow, green, blue, purple}

- Find
*A*⋃*B* - Find
*A*⋂*B* - Find
*A*⋂^{c}*C*

- The union contains all the elements in either set:
*A*⋃*B*= {red, green, blue, yellow, orange} Notice we only list red once. - The intersection contains all the elements in both sets:
*A*⋂*B*= {red} - Here we’re looking for all the elements that are
*not*in set*A*and are also in*C*.⋂*A*^{c}*C*= {orange, yellow, purple}

- If we were discussing searching for books, the universal set might be all the books in the library.
- If we were grouping your Facebook friends, the universal set would be all your Facebook friends.
- If you were working with sets of numbers, the universal set might be all whole numbers, all integers, or all real numbers

- Find (
*H*⋂*F*) ⋃*W* - Find
*H*⋂ (*F*⋃*W*) - Find (
*H*⋂*F*)⋂^{c}*W*

- We start with the intersection:
*H*⋂*F*= {dog, rabbit}. Now we union that result with*W*: (*H*⋂*F*) ⋃*W*= {dog, duck, rabbit, deer, frog, mouse} - We start with the union:
*F*⋃*W*= {dog, cow, rabbit, duck, pig, deer, frog, mouse}. Now we intersect that result with*H*:*H*⋂ (*F*⋃*W*) = {dog, rabbit, mouse} - We start with the intersection:
*H*⋂*F*= {dog, rabbit}. Now we want to find the elements of*W*that are*not*in*H*⋂*F.*(*H*⋂*F)*⋂^{c}*W*= {duck, deer, frog, mouse}

- Tea only
- Coffee only
- Both coffee and tea

- Have used both

n(*F* ⋃ *T*) = n(*F*) + n(*T*) – n(*F* ⋂ *T*)
n(*F* ⋃ *T*) = 70% + 40% – 20% = 90%

21 were taking a SS course | 26 were taking a HM course |

19 were taking a NS course | 9 were taking SS and HM |

7 were taking SS and NS | 10 were taking HM and NS |

3 were taking all three | 7 were taking none |

43 believed in UFOs | 44 believed in ghosts |

25 believed in Bigfoot | 10 believed in UFOs and ghosts |

8 believed in ghosts and Bigfoot | 5 believed in UFOs and Bigfoot |

2 believed in all three |

- Circle X has area 9π square units.

- Circle X has radius equal to 3.
- If any circle has radius
*r*, then its area is π*r*^{2}square units. - Circle X has area 9π square units.

If a circle has radius *r*, then its area is π*r*^{2} square units.

Every even number is divisible by 2.

[latex]2\text{ }{\in}\text{ }\mathbb{Z}[/latex] (2 is an element of the set of integers (or more simply, 2 is an integer).)

[latex]\sqrt{2}\text{ }{\notin}\text{ }\mathbb{Z}[/latex] (The square root of 2 is not an integer.)

[latex]\mathbb{N}\text{ }{\subseteq}\text{ }\mathbb{Z}[/latex] (The set of natural numbers is a subset of the set of integers.)

The set {0,1,2} has three elements.

Some right triangles are isosceles.

All right triangles are isosceles.

5 = 2

[latex]\sqrt{2}\text{ }{\notin}\text{ }\mathbb{R}[/latex] (The square root of 2 is not a real number.)

[latex]\mathbb{Z}\subseteq\mathbb{N}[/latex] (The set of integers is a subset of the set of natural numbers.)

[latex]{0,1,2}\cap\mathbb{N}=\varnothing[/latex] (The intersection of the set {0,1,2} and the natural numbers is the empty set.)

NOT Statements | Statements |
---|---|

Add 5 to both sides. | Adding 5 to both sides of x − 5 = 37 gives x = 42. |

[latex]\mathbb{Z}[/latex] (The set of integers) | [latex]42\text{ }{\in}\text{ }\mathbb{Z}[/latex] (42 is an element of the set of integers.) |

42 | 42 is not a number. |

What is the solution of 2x = 84? |
The solution of 2x = 84 is 42. |

*P* : For every integer *n* > 1, the number 2* ^{n}* − 1 is prime.

*P* : If an integer *x* is a multiple of 6, then *x* is even.

*P*(*x*) : If an integer *x* is a multiple of 6, then *x* is even.

*Q*(*x*) : The integer *x* is even.

*R*(*f*, *g*) : The function *f* is the derivative of the function *g*.

*P* : The solutions of the equation [latex]\displaystyle{a}x^2+bx+c=0\text{ are }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex].
*Q* : If a right triangle has legs of lengths *a* and *b* and hypotenuse of length [latex]a^2+b^2={c}^2[/latex]..

R : For all numbers *a*, *b*, *c*, *n* ∈ [latex]\mathbb{N}[/latex] with *n* > 2, it is the case that [latex]a^n+b^n\neq{c}^n[/latex].

*S* : Every even integer greater than 2 is a sum of two prime numbers.

*R*_{1} : The number 2 is even **and** the number 3 is odd.

*P* : The number 2 is even.
*Q* : The number 3 is odd.

*R*_{2} : The number 1 is even **and** the number 3 is odd.
*R*_{3} : The number 2 is even **and** the number 4 is odd.
*R*_{4} : The number 3 is even **and** the number 2 is odd.

P |
Q |
P ∧ Q |
---|---|---|

T | T | T |

T | F | F |

F | T | F |

F | F | F |

*S*_{1} : The number 2 is even or the number 3 is odd.
*S*_{2} : The number 1 is even or the number 3 is odd.
*S*_{3} : The number 2 is even or the number 4 is odd.
*S*_{4} : The number 3 is even or the number 2 is odd.

P |
Q |
P ∨ Q |
---|---|---|

T | T | T |

T | F | T |

F | T | T |

F | F | F |

*P* **or** *Q*, **but not both.**

**Either** *P* **or** *Q*.

Pay your tuition **or** you will be withdrawn from school, **but not both**.
**Either** you pay your tuition **or** you will be withdrawn from school.

The number 2 is even.

This statement is true. Now change it by inserting the words “It is not true that” at the beginning:**It is not true that** the number 2 is even.

For another example, starting with the false statement “[latex]2\in\varnothing[/latex]” we get the true statement “It is not true that [latex]2\in\varnothing[/latex].”

We use the symbol ∼ to stand for the words “It’s not true that,” so ∼P |
∼ P |
---|---|

T | F |

F | T |

*P* : The number 2 is even.

∼ *P* : It’s not true that the number 2 is even.
∼ *P* : It is false that the number 2 is even.
∼ *P* : The number 2 is not even.

*R* : If the integer a is a multiple of 6, then a is divisible by 2.

P : The integer a is a multiple of 6. Q : The integer a is divisible by 2. R : If P, then Q.

In general, given any two statementsP |
Q |
P ⇒ Q |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

**If** you pass the final exam, **then** you will pass the course.

(You pass the exam) ⇒ (You pass the course).

Under what circumstances did she lie? There are four possible scenarios, depending on whether or not you passed the exam and whether or not you passed the course. These scenarios are tallied in the following table.You pass exam | You pass course | (You pass exam) ⇒ (You pass course) |
---|---|---|

T | T | T |

T | F | F |

F | T | T |

F | F | T |

(You pass the exam) ⇒ (You pass the course)

This means that your passing the exam is a sufficient (though perhaps not necessary) condition for your passing the course. Thus your professor might just as well have phrased her promise in one of the following ways.Passing the exam is a sufficient condition for passing the course. For you to pass the course, it is sufficient that you pass the exam.

However, when we want to say “(a is a multiple of 6) ⇒ (a is divisible by 2), (a is divisible by 2) ⇒ (a is a multiple of 6).

The first statement asserts that if a is a multiple of 6 then a is divisible by 2. This is clearly true, for any multiple of 6 is even and therefore divisible by 2. The second statement asserts that if a is divisible by 2 then it is a multiple of 6. This is not necessarily true, for a = 4 (for instance) is divisible by 2, yet not a multiple of 6. Therefore the meanings ofHowever, the **contrapositive** of *P* ⇒ *Q*, ~*Q* ⇒ ~*P*, is equivalent to *P* ⇒ *Q*. Similarly, the **inverse** of *P* ⇒ *Q*, which is ~*P* ⇒ ~*Q*, is equivalent to the converse *Q* ⇒ *P*. In "Truth Tables for Statements," we will learn how to show these equivalences using a truth table.

But sometimes, if (*a* is even) ⇒ (*a* is divisible by 2),
(*a* is divisible by 2) ⇒ (*a* is even).

(*a* is even) ⇔ (*a* is divisible by 2),

P | Q | P ⇔ Q |
---|---|---|

T | T | T |

T | F | F |

F | T | F |

F | F | T |

(*P* ∨ *Q*)∧ ∼ (*P* ∧ *Q*),

*P or Q is true, and it is not the case that both P and Q are true.*

P |
Q |
(P ∨ Q) |
(P ∧ Q) |
∼(P ∧ Q) |
(P ∨ Q)∧ ∼(P ∧ Q) |
---|---|---|---|---|---|

T | T | T | T | F | F |

T | F | T | F | T | T |

F | T | T | F | T | T |

F | F | F | F | T | F |

The product *xy* equals zero if and only if *x* = 0 or *y* = 0.

P |
Q |
R |
Q ∨ R |
P ⇔ (Q ∨ R) |
---|---|---|---|---|

T | T | T | T | T |

T | T | F | T | T |

T | F | T | T | T |

T | F | F | F | F |

F | T | T | T | F |

F | T | F | T | F |

F | F | T | T | F |

F | F | F | F | T |

You pass the class if and only if you get an “A” on the final or you get a “B” on the final.

This promise has the formP |
Q |
∼P |
∼Q |
(P ∧ Q) |
(∼P ∧ ∼Q) |
(P ∧ Q)∨(∼P ∧ ∼Q) |
P ⇔ Q |
---|---|---|---|---|---|---|---|

T | T | F | F | T | F | T | T |

T | F | F | T | F | F | F | F |

F | T | T | F | F | F | F | F |

F | F | T | T | F | T | T | T |

*P* ⇔ *Q* = (*P* ∧ *Q*)∨(∼*P *∧ ∼*Q*)

P |
Q |
∼P |
∼Q |
(∼Q) ⇒ (∼P) |
P ⇒ Q |
---|---|---|---|---|---|

T | T | F | F | T | T |

T | F | F | T | F | F |

F | T | T | F | T | T |

F | F | T | T | T | T |

- ∼(
*P*∧*Q*) = (∼*P*)∨(∼*Q*) - ∼(
*P*∨*Q*) = (∼*P*)∧(∼*Q*)

P |
Q |
~P |
~Q |
P ∧ Q |
∼(P ∧ Q) |
(∼P)∨(∼Q) |
---|---|---|---|---|---|---|

T | T | F | F | T | F | F |

T | F | F | T | F | T | T |

F | T | T | F | F | T | T |

F | F | T | T | F | T | T |

∼ (P ∧Q) = (∼ P)∨(∼ Q) ∼ (P ∨Q) = (∼ P)∧(∼ Q)

(from "Logical Equivalence") can be viewed as rules that tell us how to negate the statements P ∧Q and P ∨Q. Here are some examples that illustrate how DeMorgan’s laws are used to negate statements involving “and” or “or.”*R* : You can solve it by factoring or with the quadratic formula.

∼(*P* ∨ *Q*) = (∼*P*)∧(∼*Q*).

∼*R* : You can’t solve it by factoring and you can’t solve it with the quadratic formula.

*R* : The numbers x and y are both odd.

~[(*x* is odd) ∧ (*y* is odd)] = ∼(*x* is odd) ∨ ∼(*y* is odd)
(*x* is odd) ∧ (*y* is odd) = (*x* is even) ∨ (*y* is even).

∼ R : The number x is even or the number y is even. ∼ R : At least one of x and y is even.

It is not the case that *P*(*x*) is true for all natural numbers *x*.

∼ (∀*x* ∈ *S*, *P*(*x*)) = ∃ *x* ∈ *S*, ∼*P*(*x*)
∼ (∃ *x* ∈ *S*, *P*(*x*)) = ∀*x* ∈ *S*, ∼*P*(*x*)

The first two statements in each case are called “premises” and the final statement is the “conclusion.” We combine premises with ∧ (“and”). The premises together imply the conclusion. Thus, the first argument would have (( *P* ⇒ *Q*) ∧ *P*) ⇒ *Q* as its symbolic statement.

S |
C |
S or C |
---|---|---|

T | T | T |

T | F | T |

F | T | T |

F | F | F |

A | B | A ⋀ B |
---|---|---|

T | T | T |

T | F | F |

F | T | F |

F | F | F |

A | B | A ⋁ B |
---|---|---|

T | T | T |

T | F | T |

F | T | T |

F | F | F |

A | ~A |
---|---|

T | F |

F | T |

A |
B |
C |
B ⋁ C |

T | T | T | T |

T | T | F | T |

T | F | T | T |

T | F | F | F |

F | T | T | T |

F | T | F | T |

F | F | T | T |

F | F | F | F |

A |
B |
C |
B ⋁ C |
~(B ⋁ C) |

T | T | T | T | F |

T | T | F | T | F |

T | F | T | T | F |

T | F | F | F | T |

F | T | T | T | F |

F | T | F | T | F |

F | F | T | T | F |

F | F | F | F | T |

A |
B |
C |
B ⋁ C |
~(B ⋁ C) |
A ⋀ ~(B ⋁ C) |

T | T | T | T | F | F |

T | T | F | T | F | F |

T | F | T | T | F | F |

T | F | F | F | T | T |

F | T | T | T | F | F |

F | T | F | T | F | F |

F | F | T | T | F | F |

F | F | F | F | T | F |

- You upload the picture and keep your job
- You upload the picture and lose your job
- You don’t upload the picture and keep your job
- You don’t upload the picture and lose your job

p |
q |
p → q |

T | T | T |

T | F | F |

F | T | T |

F | F | T |

m |
p |
~p |
m ⋀ ~p |

T | T | F | F |

T | F | T | T |

F | T | F | F |

F | F | T | F |

m |
p |
~p |
m ⋀ ~p |
r |
(m ⋀ ~p) → r |

T | T | F | F | T | T |

T | F | T | T | T | T |

F | T | F | F | T | T |

F | F | T | F | T | T |

T | T | F | F | F | T |

T | F | T | T | F | F |

F | T | F | F | F | T |

F | F | T | F | F | T |

Implication | Converse | Inverse | Contrapositive | ||
---|---|---|---|---|---|

p |
q |
p → q |
q → p |
~p → ~q |
~q → ~p |

T | T | T | T | T | T |

T | F | F | T | T | F |

F | T | T | F | F | T |

F | F | T | T | T | T |

I forgot my purse last week I forgot my purse today

The conclusion is:I always forget my purse

Notice that the premises are specific situations, while the conclusion is a general statement. In this case, this is a fairly weak argument, since it is based on only two instances.All cats are mammals A tiger is a cat

The conclusion is:A tiger is a mammal

Both the premises are true. To see that the premises must logically lead to the conclusion, one approach would be use a Venn diagram. From the first premise, we can conclude that the set of cats is a subset of the set of mammals. From the second premise, we are told that a tiger lies within the set of cats. From that, we can see in the Venn diagram that the tiger also lies inside the set of mammals, so the conclusion is valid.- Draw a Venn diagram based on the premises of the argument
- If the premises are insufficient to determine what determine the location of an element, indicate that.
- The argument is valid if it is clear that the conclusion must be true

Premise: All firefighters know CPR Premise: Jill knows CPR Conclusion: Jill is a firefighter

From the first premise, we know that firefighters all lie inside the set of those who know CPR. From the second premise, we know that Jill is a member of that larger set, but we do not have enough information to know if she also is a member of the smaller subset that is firefighters. Since the conclusion does not necessarily follow from the premises, this is an invalid argument, regardless of whether Jill actually is a firefighter.Premise: If you live in Seattle, you live in Washington. Premise: Marcus does not live in Seattle Conclusion: Marcus does not live in Washington

From the first premise, we know that the set of people who live in Seattle is inside the set of those who live in Washington. From the second premise, we know that Marcus does not lie in the Seattle set, but we have insufficient information to know whether or not Marcus lives in Washington or not. This is an invalid argument.Premise: If you bought bread, then you went to the store Premise: You bought bread Conclusion: You went to the store

While this example is hopefully fairly obviously a valid argument, we can analyze it using a truth table by representing each of the premises symbolically. We can then look at the implication that the premises together imply the conclusion. If the truth table is a tautology (always true), then the argument is valid. We’ll get B represent “you bought bread” and S represent “you went to the store”. Then the argument becomes:Premise: *B* → *S
*Premise: *B
*Conclusion: *S*

B |
S |
B → S |
(B→S) ⋀ B |
[(B→S) ⋀ B] → S |

T | T | T | T | T |

T | F | F | F | T |

F | T | T | F | T |

F | F | T | F | T |

- Represent each of the premises symbolically
- Create a conditional statement, joining all the premises with and to form the antecedent, and using the conclusion as the consequent.
- Create a truth table for that statement. If it is always true, then the argument is valid.

Premise: If I go to the mall, then I’ll buy new jeans Premise: If I buy new jeans, I’ll buy a shirt to go with it Conclusion: If I got to the mall, I’ll buy a shirt.

LetPremise: *M* → *J
*Premise: *J* → *S
*Conclusion: *M* → *S*

M |
J |
S |
M → J |
J → S |
(M→J) ⋀ (J→S) |
M → S |
[(M→J) ⋀ (J→S)] → (M→S) |

T | T | T | T | T | T | T | T |

T | T | F | T | F | F | F | T |

T | F | T | F | T | F | T | T |

T | F | F | F | T | F | F | T |

F | T | T | T | T | T | T | T |

F | T | F | T | F | F | T | T |

F | F | T | T | T | T | T | T |

F | F | F | T | T | T | T | T |

Base 2 number | Base 10 equivalent | Base 8 number |
---|---|---|

1000 | 8 | 10 = 1 × 8 + 0 × 1 |

1001 | 9 | 11 = 1 × 8 + 1 × 1 |

1010 | 10 | 12 = 1 × 8 + 2 × 1 |

… | … | … |

111100 | 60 | 74 = 7 × 8 + 4 × 1 |

111101 | 61 | 75 = 7 × 8 + 5 × 1 |

111110 | 62 | 76 = 7 × 8 + 6 × 1 |

111111 | 63 | 77 = 7 × 8 + 7 × 1 |

- Convert the number 6157
_{8}to base 2. We split each digit in base 8 to three digits in base 2, using the three digit base 2 equivalent, so 6_{8}= 110_{2}, 1_{8}= 001_{2}, etc. - Convert the number 10111011001010
_{2}to base 8. Split this number into sets of three,**starting with the right-most digit**, then convert each set of three to its equivalent in base 8.

- At the first logical geometric “proofs” traditionally credited to Thales of Miletus (600 BCE).
- With the formulation of methods of measurement made by the Egyptians and Mesopotamians/Babylonians.
- Where prehistoric peoples made efforts to organize the concepts of size, shape, and number.
- In pre-human times in the very simple number sense and pattern recognition that can be displayed by certain animals, birds, etc.
- Even before that in the amazing relationships of numbers and shapes found in plants.
- With the spiral nebulae, the natural course of planets, and other universe phenomena.

- Eleven comes from “ein lifon,” meaning “one left over.”
- Twelve comes from “twe lif,” meaning “two left over.”
- Thirteen comes from “Three and ten” as do fourteen through nineteen.
- Twenty appears to come from “twe-tig” which means “two tens.”
- Hundred probably comes from a term meaning “ten times.”

Using the [Indian] numerals, multiplication and division are carried out. Each numeral is written in one stroke. When a number is counted to ten, it is advanced into the higher place. In each vacant place a dot is always put. Thus the numeral is always denoted in each place. Accordingly there can be no error in determining the place. With the numerals, calculations is easy.[footnote]Ibid, page 232.[/footnote]

5,000,000 | = 5 × 1,000,000 | = 5 × 10^{6} |
Five million |

+700,000 | = 7 × 100,000 | = 7 × 10^{5} |
Seven hundred thousand |

+80,000 | = 8 × 10,000 | = 8 × 10^{4} |
Eighty thousand |

+3,000 | = 3 × 1000 | = 3 × 10^{3} |
Three thousand |

+200 | = 2 × 100 | = 2 × 10^{2} |
Two hundred |

+10 | = 1 × 10 | = 1 × 10^{1} |
Ten |

+6 | = 6 × 1 | = 6 × 10^{0} |
Six |

5,783,216 | Five million, seven hundred eighty-three thousand, two hundred sixteen |

Base 5 | This column coverts to base-ten | In Base-Ten | |

3 × 5^{4} |
= 3 × 625 | = 1875 | |

+ | 0 × 5^{3} |
= 0 × 125 | = 0 |

+ | 4 × 5^{2} |
= 4 × 25 | = 100 |

+ | 1 × 5^{1} |
= 1 × 5 | = 5 |

+ | 2 × 5^{0} |
= 2 × 1 | = 2 |

Total | 1982 |

Base 7 | Convert | Base 10 | |

= 6 × 7^{3} |
= 6 × 343 | = 2058 | |

+ | = 2 × 7^{2} |
= 2 × 49 | = 98 |

+ | = 3 × 7 | = 3 × 7 | = 21 |

+ | = 4 × 1 | = 4 × 1 | = 4 |

Total | 2181 |

12 = (2 × 5) + (2 × 1)

Hence, we have two fives and 2 ones. Hence, in base-five we would write twelve as 2269 = (2 × 25) + (3 × 5) + (4 × 1)

Here, we have two twenty-fives, 3 fives, and 4 ones. Hence, in base five we have 234. Thus, 697^{0} = 1
7^{1} = 7
7^{2} = 49
7^{3} = 343
Etc…

- Find the highest power of the base
*b*that will divide into the given number at least once and then divide. - Write down the whole number part, then use the remainder from division in the next step.
- Repeat step two, dividing by the next highest power of the base
*b*, writing down the whole number part (including 0), and using the remainder in the next step. - Continue until the remainder is smaller than the base. This last remainder will be in the “ones” place.
- Collect all your whole number parts to get your number in base
*b*notation.

5^{0} = 1
5^{1} = 5
5^{2} = 25
5^{3} = 125
5^{4} = 625
Etc…

348 ÷ 125 = 2 with remainder 98

We write down the whole part, 2, and continue with the remainder. There are 98 left over, so we see how many 25s (the next smallest power of five) there are in the remainder:98 ÷ 25 = 3 with remainder 23

We write down the whole part, 2, and continue with the remainder. There are 23 left over, so we look at the next place, the 5s:23 ÷ 5 = 4 with remainder 3

This leaves us with 3, which is less than our base, so this number will be in the “ones” place. We are ready to assemble our base-five number:348 = (2 × 5^{3}) + (3 × 5^{2}) + (4 × 5^{1}) + (3 × 1)

7^{0} = 1
7^{1} = 7
7^{2} = 49
7^{3} = 343
7^{4} = 2401
7^{5} = 16807
Etc…

4,509 ÷ 7^{4} = 1 R 21082108 ÷ 7^{3} = 6 R 50
50 ÷ 7^{2} = 1 R 1
1 ÷ 7^{1} = 1
4,509_{10} = 16101_{7}.

4863 ÷ 1000 = 4.863

This says that there are four thousands in 4863 (obviously). However, it also says that there are 0.863 thousands in 4863. This fractional part is our remainder and will be converted to lower powers of our base (10). If we take that decimal and multiply by 10 (since that’s the base we’re in) we get the following:0.863 × 10 = 8.63

Why multiply by 10 at this point? We need to recognize here that 0.863 thousands is the same as 8.63 hundreds. Think about that until it sinks in.(0.863)(1000) = 863 (8.63)(100) = 863

These two statements are equivalent. So, what we are really doing here by multiplying by 10 is rephrasing or converting from one place (thousands) to the next place down (hundreds).0.863 × 10 ⇒ 8.63 (Parts of Thousands) × 10 ⇒ Hundreds

What we have now is 8 hundreds and a remainder of 0.63 hundreds, which is the same as 6.3 tens. We can do this again with the 0.63 that remains after this first step.0.63 × 10 ⇒ 6.3 Hundreds × 10 ⇒ Tens

So we have six tens and 0.3 tens, which is the same as 3 ones, our last place value. Now here’s the punch line. Let’s put all of the together in one place: Converting from Base 10 to Base- Find the highest power of the base
*b*that will divide into the given number at least once and then divide. - Keep the whole number part, and multiply the fractional part by the base
*b*. - Repeat step two, keeping the whole number part (including 0), carrying the fractional part to the next step until only a whole number result is obtained.
- Collect all your whole number parts to get your number in base
*b*notation.

5^{0} = 1
5^{1} = 5
5^{2} = 25
5^{3} = 125
5^{4} = 625
Etc…

3007 ÷ 625 = ④.8112 0.8112 × 5 = ④.056 0.056 × 5 = ⓪.28 0.28 × 5 = ①0.4 0.4 × 5 = ②0.0

This gives us that 30077^{0} = 1
7^{1} = 7
7^{2} = 49
7^{3} = 343
7^{4} = 2401
7^{5} = 16807
etc…

63201 ÷ 7^{5} = 3.760397453

63201 ÷ 7^{5} = ③.760397453
0.760397453 × 7 =
⑤.322782174
0.322782174 × 7 = ②.259475219
0.259475219 × 7 = ①.816326531
0.816326531 × 7 = ⑤
.714285714
0.714285714 × 7 =
⑤.000000000

Powers | Base-Ten Value | Place Name |

20^{7} |
12,800,000,000 | Hablat |

20^{6} |
64,000,000 | Alau |

20^{5} |
3,200,000 | Kinchil |

20^{4} |
160,000 | Cabal |

20^{3} |
8,000 | Pic |

20^{2} |
400 | Bak |

20^{1} |
20 | Kal |

20^{0} |
1 | Hun |

3575 ÷ 400 = 8.9375 0.9375 × 20 = 18.75 0.75 × 20 = 15.0

This means that 3575- favorable outcomes = 1(red)
- unfavorable outcomes = 2(blue, yellow)
- total outcomes = 3

Odds in favor of rain: 7 to 3

These odds tell you not only the odds of rain, but also the odds of not raining. If the odds in favor or rain are 7 to 3, then the odds against rain are:Odds against rain: 3 to 7

Another way of saying that is:Odds that it will NOT rain: 3 to 7

You can use this idea in many different situations. If you know the odds that something will happen, then you also know the odds that it will not happen. Use this spinner to calculate odds.- favorable outcomes = 1(4)
- unfavorable outcomes = 5(1,2,3,5,6)

- For rolling a number cube, what are the odds in favor of rolling a 2?
- For rolling a number cube, what are the odds against rolling a 2?
- For rolling a number cube, what are the odds in favor of rolling a number greater than 3?
- For rolling a number cube, what are the odds in favor rolling a number less than 5?
- For rolling a number cube, what are the odds against rolling a number less than 5?
- For rolling a number cube, what are the odds in favor of rolling an even number?
- For rolling a number cube, what are the odds against rolling an even number?

- For spinning the spinner, what are the odds in favor of the arrow landing on 10?
- For spinning the spinner, what are the odds in favor of the arrow landing on a 2 or 3?
- For spinning the spinner, what are the odds in favor of the arrow landing on 7, 8 or 9?
- For spinning the spinner, what are the odds in favor of NOT landing on an even number?
- For spinning the spinner, what are the odds of the arrow NOT landing on 10?
- For spinning the spinner, what are the odds in favor of the arrow landing on a number greater than 2?
- For spinning the spinner, what are the odds in favor of the arrow NOT landing on a number greater than 2?
- For spinning the spinner, what are the odds of the arrow not landing on a number greater than 3?

Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?

Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however.Outcome | Probability of outcome |
---|---|

$35 | [latex]\displaystyle\frac{{1}}{{38}}[/latex] |

–$1 | [latex]\displaystyle\frac{{37}}{{38}}[/latex] |

Outcome | Probability of outcome |
---|---|

$999,999 | [latex]\frac{{1}}{{12271512}}[/latex] |

$999 | [latex]\frac{{252}}{{12271512}}[/latex] |

–$1 | [latex]{1}-\frac{{253}}{{12271512}}=\frac{{12271259}}{{12271512}}[/latex] |

Outcome | Probability of outcome |
---|---|

$100,000 – $275 = $99,725 | 0.00242 |

–$275 | 1 – 0.00242 = 0.99758 |

- We roll a 1
- We roll a 5

- We roll a number bigger than 4
- We roll an even number

*P*(rolling a 1)*P*(rolling a number bigger than 4)

- There is one outcome corresponding to “rolling a 1,” so the probability is [latex]\frac{1}{6}[/latex]
- There are two outcomes bigger than a 4, so the probability is [latex]\frac{2}{6}=\frac{1}{3}[/latex]

- What is probability the minutes reading is 15?
- What is the probability the minutes reading is 15 or less?

- A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.
- The two events (1) "It will rain tomorrow in Houston" and (2) "It will rain tomorrow in Galveston” (a city near Houston).
- You draw a card from a deck, then draw a second card without replacing the first.

- The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss, so these events are independent.
- These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.
- The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.

- Has a red car
*and*got a speeding ticket - Has a red car
*or*got a speeding ticket.

Speeding Ticket | No Speeding Ticket | Total | |
---|---|---|---|

Red car | 15 | 135 | 150 |

Not red car | 45 | 470 | 515 |

Total | 60 | 605 | 665 |

- Has a speeding ticket
*given*they have a red car - Has a red car
*given*they have a speeding ticket

Speeding Ticket | No Speeding Ticket | Total | |
---|---|---|---|

Red car | 15 | 135 | 150 |

Not red car | 45 | 470 | 515 |

Total | 60 | 605 | 665 |

- Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so [latex]P(\text{ticket | red car})=\frac{15}{150}=\frac{1}{10}=0.1[/latex]
- Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so [latex]P(\text{red car | ticket})=\frac{15}{60}=\frac{1}{4}=0.25[/latex]

- Case A: you can get the Ace of Diamonds first and then a black card or
- Case B: you can get a black card first and then the Ace of Diamonds.

*P*(not pregnant | positive test result)*P*(positive test result | not pregnant)

Positive Test | Negative Test | Total | |
---|---|---|---|

Pregnant | 70 | 4 | 74 |

Not Pregnant | 5 | 14 | 19 |

Total | 75 | 18 | 93 |

- Since we know the test result was positive, we’re limited to the 75 women in the first column, of which 5 were not pregnant.
*P*(not pregnant | positive test result) = [latex]\frac{5}{75}\approx{0.067}[/latex]. - Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.
*P*(positive test result | not pregnant) = [latex]\frac{5}{19}\approx{0.263}[/latex].

- 10 mod 3
- 15 mod 5
- 27 mod 5

- Since 10 divided by 3 is 3 with remainder 1, 10 mod 3 = 1
- Since 15 divided by 5 is 3 with no remainder, 15 mod 5 = 0
- 2
^{7}= 128. 128 divide by 5 is 25 with remainder 3, so 2^{7}mod 5 = 3

- 23 mod 7
- 15 mod 7
- 2034 mod 7

- Divide
*a*by*n* - Subtract the whole part of the resulting quantity
- Multiply by
*n*to obtain the modulus

[latex]31345\div{419}=74.8090692[/latex] | Now subtract 74 to get just the decimal remainder |

[latex]74.8090692-74=0.8090692[/latex] | Multiply this by 419 to get the modulus |

[latex]0.8090692\times{419}=339[/latex] | This tells us 0.8090692 was equivalent to [latex]\frac{339}{419}[/latex] |

1) | 1351 0 (mod 7) | 1) | |

A) False | B) True | ||

2) | 11 | 4 (mod 7) | 2) |

A) False | B) True | ||

3) | 66 | 7 (mod 12) | 3) |

A) False | B) True |

4) | (6 + 5) (mod 6) | 4) | ||

A) 5 | B) 6 | C) 11 | D) 4 | |

5) | (48 + 48) (mod 50) | 5) | ||

A) 4 | B) 46 | C) 50 | D) 96 |

6) | 8 | + 10 in 12-hour clock arithmetic | 6) | ||

A) 2 | B) 8 | C) 6 | D) 0 | ||

7) | 7 | · 16 in 12-hour clock arithmetic | 7) | ||

A) 4 | B) 5 | C) 16 | D) 11 | ||

8) | 3 | + 221 in 7-day clock arithmetic | 8) | ||

A) 3 | B) 5 | C) 0 | D) 8 | ||

9) | 1400 + 1900 in the military 24-hour clock system | 9) | |||

A) 0930 | B) 12100 | C) 1900 | D) 0900 | ||

10) | 0930 + 1640 in the military 24-hour clock system | 10) | |||

A) 0310 | B) 2610 | C) 0210 | D) 2570 |

11) | 6 13 (mod 2) | 11) |

A) True | B) False | |

12) | 0 26 (mod 7) | 12) |

A) True | B) False | |

13) | 19 77 (mod 5) | 13) |

A) True | B) False | |

14) | 5 21 (mod 5) | 14) |

A) True | B) False |

1

15) 3 13 (mod 11) 15)
A) True B) False
**Perform the modular arithmetic operation.**

**Find all positive solutions for the equation.**

16) | (46 + 37)(mod 7) | 16) | ||

A) 6 | B) 7 | C) 11 | D) 5 | |

17) | (130 + 106)(mod 9) | 17) | ||

A) 10 | B) 26 | C) 2 | D) 1 | |

18) | (10 · 7)(mod 6) | 18) | ||

A) 3 | B) 6 | C) 11 | D) 4 | |

19) | [(11 + 7) · (7 + 3)](mod 7) | 19) | ||

A) 4 | B) 7 | C) 25 | D) 5 | |

20) | (49 - 25)(mod 5) | 20) | ||

A) 3 | B) 0 | C) 120 | D) 4 | |

21) | (15 - 53)(mod 4) | 21) | ||

A) 3 | B) 2 | C) 1 | D) 152 | |

22) | [(3 · 7) - 5](mod 4) | 22) | ||

A) 1 | B) 3 | C) 2 | D) 0 | |

23) | [(13 · 3) + 9](mod 8) | 23) | ||

A) 3 | B) 7 | C) 0 | D) 1 | |

24) | [(4 - 9) · 7](mod 5) | 24) | ||

A) 2 | B) 0 | C) 4 | D) 3 | |

25) | [(-5) · 6](mod 7) | 25) | ||

A) -5 | B) 5 | C) -2 | D) 1 |

26) x 4 (mod 7) | 26) | |||

A) {1, 18, 25, ...} | B) {4, 11, 18, ...} | C) {4, 8, 12, ...} | D) {11, 18, 91, ...} | |

27) | 2x 1 (mod 3) | 27) | ||

A) {2, 6, 10, 14, ...} | B) {1, 4, 7, 10, ...} | |||

C) {2, 5, 8, 11, ...} | D) None | |||

28) | 2x 8 (mod 10) | 28) | ||

A) Identity | B) {4, 9, 14, 19, 24, 29, ...} | |||

C) {4, 14, 24, ...} | D) {9, 19, 29, ...} | |||

29) | 8x 4 (mod 4) | 29) | ||

A) {4, 8, 12, ...} | B) {1, 5, 9, ...} | C) Identity | D) {2, 6, 5, ...} |

2

30) | 10x 1 (mod 10) | 30) | ||

A) {1, 10, 15, ...} | B) None | C) Identity | D) {2, 7, 12, ...} | |

31) | (2 + x) 5 (mod 4) | 31) | ||

A) {4, 6, 8, 10, 12, 14, ...} | B) {0, 2, 4, 6, 8, 10, ...} | |||

C) {3, 7, 11, 15, 19, 23, ...} | D) None |