STEP-BY-STEP LESSON · §1.1
Variables
Understand variables and distinguish “for every” from “there exists”.
Symbols
- x
- variable
- ∀
- for every
- ∃
- there exists
- ℤ / ℝ
- integers / real numbers
- ⇔
- equivalent conditions
Definitions and notation for this topic
Equality
=
An assertion that the expressions on the two sides have the same value.
Equality names the relation; equals is how we read the sign.
Equality compares values; it does not update a variable. In pseudocode x:=x+1 is assignment, whereas x=x+1 is an equation.
Separate glossary example
2 + 3 = 5
Not equal
≠
The expressions have different values.
This negates equality.
Separate glossary example
2 ≠ 3
Square
a²
The result of multiplying a number by itself.
Squaring a is different from multiplying a by 2.
Separate glossary example
a² = a × a
Multiplication
2k
A multiplication sign can be omitted between a number and a letter.
Juxtaposition denotes an ordinary product when the surrounding notation says the factors are numbers. A×B for sets denotes a Cartesian product instead.
Separate glossary example
2k = 2 × k
Variable
A letter representing an object whose value may vary.
Here x = 3 makes the equation true.
Separate glossary example
x + 2 = 5
Constant
A fixed value in the expression being considered.
The term 3 stays fixed while x may vary.
Separate glossary example
x + 3
Expression
A combination of numbers, variables, and operations representing a value.
An expression alone does not assert an equality. The statement 2x+1=7 is an equation.
Separate glossary example
2x+1
Equation
An equality involving a variable whose values are being considered or sought.
Solving it means finding all allowed values of x that make the equality true. Here x=3.
Separate glossary example
x + 2 = 5
Evaluate
Find the value of an expression using the given values of its variables.
The inputs are supplied and the requested result is a value. Solving instead asks which inputs satisfy a stated condition.
Separate glossary example
Evaluate p∨q when p=F and q=T: the value is T.
Evaluate 2x+1 at x=3: 2×3+1=7.
Solve
For an equation, find all allowed values of its unknowns that make it true.
Substitution checks the value: 3+2=5.
In this course solve can also mean finding an explicit sequence satisfying a recurrence and its initial values; read the object named after the verb.
Separate glossary example
Solve x+2=5: x=3.
Let
Introduces a named object or specifies a definition for the argument.
The sentence tells you whether an object is fixed, arbitrary or defined. Assume/suppose instead explicitly introduces a hypothesis to reason from.
Separate glossary example
Let A={1,2,3}. Let n be an arbitrary integer.
Let n ∈ ℤ.
Satisfy
Make a condition true when the object is substituted into it.
Replacing x with 3 gives the true equality 3+2=5.
Separate glossary example
The number 3 satisfies x+2=5.
Equals / is equal to
The verb phrase used to read the equality sign.
Equality names the relation, equation names an equality involving variables, and equals is the verb.
Separate glossary example
Two plus three equals five: 2+3=5.
Step by step
Step 1 / 6
Variables in a condition
In x+y=0, each variable keeps its chosen value throughout the statement. If x=3, only y=−3 satisfies this condition.
x+y=0; x=3 ⇒ y=−3
A domain specifies the possible values; a condition selects which combinations satisfy it.
Worked example
Find an integer x satisfying 3x−2=10.
- Seek a solution in ℤ, then check it against the original condition.
- Add 2 to both sides to undo subtraction: 3x=12.
- Divide both sides by 3, which is nonzero: x=4.
- 4 is an integer and 3·4−2=10, so it is an admissible solution.
Optional self-check
Must one y work for every x in ∀x∃y?
Show answer
No: after x is selected, a suitable y may be chosen.