24 Operations and Properties
Louisiana Student Standards
Standard Number |
Description of Standard |
| 1.OA.B. | Understand and apply properties of operations and the relationship between addition and subtraction. |
| 3.OA.B. | Understand properties of multiplication and the relationship between multiplication and division. |
Defining a Binary Operation
This exercise set is designed to give you an understanding of what binary operations are, and to give you a deeper understanding for the commutative, associative, and distributive properties. To do this, we’re going to define and work with some nonsense operations. First, we need to get a good grasp on what an operation is. The binary operations you are familiar with are addition, subtraction, multiplication and division. This means that you are performing a rule using two numbers. For instance, we know what to do when we see the plus sign (+), the subtraction sign (–), the multiplication sign ([latex]\times[/latex]or [latex]\cdot[/latex] ), or the division sign (÷) between two numbers. There is a specific rule that we apply.
Suppose you were asked to compute [latex]5 \perp 3[/latex]. You wouldn’t know what to do unless someone told you what [latex]\perp[/latex] meant. It’s like asking someone who has never heard of addition, or seen an addition sign (+) to compute [latex]5 + 3[/latex]. In order to do a computation, the operation used must be defined. The operations you already know about are addition, multiplication, subtraction and division. You also know how to compute with exponents, and how to compare numbers (<, = or >).
Definition – ⊥
Let’s start off by defining what [latex]\perp[/latex] means. This is just a made-up operation that we’re defining, so that you’ll have a deeper understanding of binary operations. There really isn’t such an operation as [latex]\perp[/latex] in the real world.
Define [latex]\perp[/latex] like this: [latex]m \perp n = 3m + 2n + 8[/latex].
The variables I use to define this binary operation are arbitrary. I could use any letters or symbols I want. Or I could explain how to perform the binary operation [latex]\perp[/latex], which is much more cumbersome.
This is how I could explain how to compute: To compute with [latex]\perp[/latex], multiply the number before the [latex]\perp[/latex] by 3 and add this to twice the number after the [latex]\perp[/latex], and then add 8. As you can see, it is easier to explain by using variables. Here are three more ways I could have defined [latex]\perp[/latex], without changing the meaning of [latex]\perp[/latex].
[latex]a \perp b = 3a + 2b + 8[/latex]
[latex]b \perp a = 3b + 2a + 8[/latex]
[latex]x \perp y = 3x + 2y + 8[/latex]
Notice that in each case, one multiplies the first symbol by 3, adds it to twice the second symbol, and then adds 8! It’s the same rule, but I chose different variables to explain the rule. You have to pay careful attention to what the meaning of the operation is, and follow the rule exactly. It’s like working with functions in algebra.
Example 1
Let’s compute and simplify a few problems with [latex]\perp[/latex] :
a. 5[latex]\perp[/latex]3 = 3(5) + 2(3) + 8 = 15 + 6 + 8 = 29
b. 3[latex]\perp[/latex]5 = 3(3) + 2(5) + 8 = 9 + 10 + 8 = 27
c. r[latex]\perp[/latex]s = 3r + 2s + 8
Key Takeaway – Binary operations
The rule for the operation doesn’t necessarily depend on both of the variables used and, in fact, may not depend on either of them. Let’s define several new operations; some computations use only one of the variables, some use neither. Suppose there are eight new binary operations in addition to § that are defined as follows:
| [latex]\perp[/latex] | [latex]a \perp b = 3a + 2b + 8[/latex] |
| [latex]*[/latex] | [latex]a * b = a + 2b[/latex] |
| [latex]◊[/latex] | [latex]a ◊ b = a^2 + b^2[/latex] |
| [latex]∇[/latex] | [latex]a ∇ b = 2[/latex] (Notice the answer doesn’t depend on [latex]a[/latex] or [latex]b[/latex]) |
| [latex]λ[/latex] | [latex]a λ b = 3ab[/latex] |
| [latex]\dagger[/latex] | [latex]a \dagger b[/latex] = the smaller value of [latex]a[/latex] or [latex]b[/latex] |
| [latex]@[/latex] | [latex]a @ b = 2b[/latex] (Notice the answer doesn’t depend on [latex]a[/latex]) |
| [latex]Θ[/latex] | [latex]a Θ b = 2a + 2b[/latex] |
| [latex]\otimes[/latex] | [latex]a \otimes b = a^2 + b[/latex] |
Remember that a and b are just “dummy” variables. Any variables could have been used to define the above functions. The operation, [latex]*[/latex], could have been defined like this: [latex]m * n = m + 2n[/latex]. The meaning of the definition is exactly the same. To apply the definition of [latex]*[/latex] to get the answer, it says to take the first number and add it to twice the second number.
Here are a several examples for you to study before going on.
| 6 [latex]\perp[/latex] 7 = 3(6)+2(7)+8 = 18+14+8 = 40 | 2 λ 5 = 3(2)(5) = 30 |
| 7 [latex]\perp[/latex] 6 = 3(7)+2(6)+8 = 21+12+8= 41 | r λ s = 3rs |
| 4 [latex]\perp[/latex] 4 = 3(4)+2(4)+8 = 12+8+8 = 28 | 4 [latex]\dagger[/latex] 7 = 4 |
| v [latex]\perp[/latex] z = 3v + 2z + 8 | 5 [latex]\dagger[/latex] 2 = 2 |
| 5 * 3 = 5 + 2(3) = 5 + 6 = 11 | 7 [latex]\dagger[/latex] 4 = 4 |
| 4 * 7 = 4 + 2(7) = 4 + 14 = 18 | c [latex]\dagger[/latex] d = the smaller value of c or d |
| 3 * 5 = 3 + 2(5) = 3 + 10 = 13 | 5 @ 6 = 2(6) = 12 |
| v * z = z + 2z | 7 @ 3 = 2(3) = 6 |
| 5 ◊ 3 = 52 + 32 = 25 + 9 = 34 | 6 @ 5 = 2(5) = 10 |
| 4 ◊ 3 = 42 + 32 = 16 + 9 = 25 | p @ q = 2q |
| 3 ◊ 5 = 32 + 52 = 9 + 25 = 34 | 6 Θ 3 = 2(6) + 2(3) = 12 + 6 = 18 |
| v ◊ z = v2 + z2 | 5 Θ 8 = 2(5) + 2(8) = 10 + 16 = 26 |
| 5 ∇ 3 = 2 | 3 Θ 6 = 2(3) + 2(6) = 6 + 12 = 18 |
| 8 ∇ 7 = 2 | h Θ k = 2h + 2k |
| (junk) ∇ (stuff) = 2 | 4 [latex]\otimes[/latex] 7 = 42 + 7 = 16 + 7 = 23 |
| w ∇ q = 2 | 6 [latex]\otimes[/latex] 9 = 62 + 9 = 36 + 9 = 45 |
| 5 λ 2 = 3(5)(2) = 30 | 7 [latex]\otimes[/latex] 4 = 72 + 4 = 49 + 4 = 53 |
| 4 λ 7 = 3(4)(7) = 84 | z [latex]\otimes[/latex] n = z2 + n |
Exercise 1
Compute and simplify the following.
The Commutative Property
Key Takeaway – The Commutative Property
An operation, [latex]\bowtie[/latex], is commutative if for any two values, X and Y, X [latex]\bowtie[/latex] Y = Y [latex]\bowtie[/latex] X.
The operation [latex]\bowtie[/latex] is just a “dummy” operation and X and Y are dummy variables. For a particular operation to be commutative, the equation must always be true no matter what values are used for X and Y.
For instance, the operation [latex]*[/latex] is commutative only if [latex]m * n = n * m[/latex] is always true no matter what values are put in for m or n.
To show that an operation is not commutative, all you need to do is provide a counterexample (with particular values) that shows the equation is not true for at least those particular values. To prove an operation is commutative is typically more involved because you must prove it is always true no matter what values you use. You would have to switch the order of the original values (a and b, or X and Y, etc.), and show algebraically that both expressions simplify to the same thing.
From my examples, after defining the operations and the problems you worked in exercise 1, it should be clear which of the operations are not commutative.
Example 2
Let [latex]@[/latex] be defined as follows: [latex]m @ n = 2n[/latex]. Is [latex]@[/latex] commutative?
Solution:
If @ is commutative, then m @ n = n @ m for all values m and n.
But, 5 @ 6 = 12 and 6 @ 5 = 10.
Therefore, @ is not commutative since 5 @ 6 [latex]\neq[/latex] 6 @ 5.
Example 3
We will use a new operation [latex]\Omega[/latex] on the next example. Let [latex]\Omega[/latex] be defined as follows: [latex]m \Omega n = 2mn[/latex]. Is [latex]\Omega[/latex] commutative?
Solution:
If [latex]\Omega[/latex] is commutative, then m [latex]\Omega[/latex] n = n [latex]\Omega[/latex] m for all values m and n. First, I’d try some numbers in for a and b to see if I might come up with a counterexample.
5 [latex]\Omega[/latex] 6 = 2(5)(6) = 60
6 [latex]\Omega[/latex] 5 = 2(6)(5) = 60
No counterexample here. So, I use algebra to prove that m [latex]\Omega[/latex] n = n [latex]\Omega[/latex] m.
Since m [latex]\Omega[/latex] n = 2mn, and n [latex]\Omega[/latex] m = 2nm, the question is: Does 2mn = 2nm?
Yes, it does! Since m [latex]\Omega[/latex] n = n [latex]\Omega[/latex] m for all m and n, then [latex]\Omega[/latex] is commutative.
Exercise 2
For each operation listed, determine whether it is commutative or not. If it is not commutative, give a counterexample, like I did for [latex]@[/latex]. If it is commutative, prove it is commutative, like I did for [latex]\Omega[/latex] above. Begin each problem by stating what equation must be true if the operation listed is commutative
a. [latex]∇[/latex] is defined like this: [latex]m ∇ n = 2[/latex].
Write the general equation that is true if [latex]∇[/latex] is commutative:
Solution
m ∇ n = 2 = n ∇ m
If you answered yes, prove [latex]∇[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
3 ∇ 4 = 2 and 4 ∇ 3 = 2 (The answer is 2 regardless of the values for m and n.)
b. [latex]λ[/latex] is defined like this: [latex]m λ n = 3mn[/latex].
Write the general equation that is true if [latex]λ[/latex] is commutative:
Solution
3mn = 3nm
If you answered yes, prove [latex]λ[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
3(2)(4) = 3(4)(2)
24 = 24
c. [latex]\dagger[/latex] is defined: [latex]m \dagger n[/latex] = the smaller value of m or n.
Write the general equation that is true if [latex]\dagger[/latex] is commutative:
Solution
m [latex]\dagger[/latex] n = smaller value = n [latex]\dagger[/latex] m
If you answered yes, provide an example. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
2 [latex]\dagger[/latex] 3 = 2 and 3 [latex]\dagger[/latex] 2 = 2 (The answer is always the smaller value.)
d. [latex]Θ[/latex] is defined like this: [latex]m Θ n = 2m + 2n[/latex].
Write the general equation that is true if [latex]Θ[/latex] is commutative:
Solution
2m + 2n = 2n + 2m
If you answered yes, prove [latex]Θ[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
2(3) + 2(4) = 2(4) + 2(3)
6 + 8 = 8 + 6
14 = 14
e. [latex]\otimes[/latex] is defined like this: [latex]m \otimes n = m^2+ n[/latex].
Write the general equation that is true if [latex]\otimes[/latex] is commutative:
Solution
[latex]\otimes[/latex] is not commutative
If you answered yes, prove [latex]\otimes[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
22 + 3 ≠ 32 + 2
4 + 3 ≠ 9 + 2
7 ≠ 11
f. [latex]◊[/latex] is defined like this: [latex]m ◊ n = m^2 + n^2[/latex].
Write the general equation that is true if [latex]◊[/latex] is commutative:
Solution
m2 + n2 = n2 + m2
If you answered yes, prove [latex]◊[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
22 + 32 = 32 + 22
4 + 9 = 9 + 4
13 = 13
g. [latex]\perp[/latex] is defined like this: [latex]m \perp n = 3m + 2n + 8[/latex].
Write the general equation that is true if [latex]\perp[/latex] is commutative:
Solution
[latex]\perp[/latex] is not commutative
If you answered yes, prove [latex]\perp[/latex] is commutative. If you answered no, provide a counterexample to illustrate it is not commutative.
Solution
3(2) + 2(3) + 8 ≠ 3(3) + 2(2) + 8
6 + 6 + 8 ≠ 9 + 4 + 8
20 ≠ 21
Before going on to determining whether an operation is associative or distributive, we should compute a few more problems, which are a bit more involved. Make sure you follow the order of operations as you work through these next few problems. Look at the examples first.
Example 4
Solve the following problems:
a. Solve (2 λ 3) λ 4
Solution:
First do 2 λ 3 = 3(2)(3) = 18
Now do 18 λ 4 = 3(18)(4) = 216
so, (2 λ 3) λ 4 = 216
a. Solve 4 ◊ (3 ◊ 2)
Solution:
First do 3 ◊ 2 = 32 + 22 = 9 + 4 = 13
Now do 4 ◊ 13 = 42 + 132 = 16 + 169 = 185
so, 4 ◊ (3 ◊ 2) = 185
Exercise 3
a. (3 * 5) * 2
Solution
(3 * 5) * 2 = [ 3 + 2( 5 ) ] * 2
= [3 + 10] * 2
= 13 * 2
= 13 + 2( 2 )
= 13 + 4
= 17
b. (3 @ 5) @ 2
Solution
(3 @ 5) @ 2 = [ 2( 5 ) ] @ 2
= 10 @ 2
= 2( 2 )
= 4
c. (3 ∇ 5) ∇ 7
Solution
(3 ∇ 5) ∇ 7 = 2 ∇ 7
= 2
d. (3 λ 4) λ 2
Solution
(3 λ 4) λ 2 =[ 3( 3 )( 4 )] λ 2
= 36 λ 2
= 3( 36 )( 2 )
= 216
e. (3 Θ 5) Θ 2
Solution
(3 Θ 5) Θ 2= [2( 3 ) + 2( 5 )] Θ 2
= [6 + 10] Θ 2
= 16 Θ 2
= 2( 16 ) + 2( 2 )
= 32 + 4
= 36
f. (3 [latex]\otimes[/latex] 2) [latex]\otimes[/latex] 4
Solution
(3 [latex]\otimes[/latex] 2) [latex]\otimes[/latex] 4 = [ 32 + 2 ] [latex]\otimes[/latex] 4
= [9 + 2] [latex]\otimes[/latex] 4
= 11 [latex]\otimes[/latex] 4
= 112 + 4
= 121 + 4
= 125
Exercise 4
Compute the following, using the definitions for the operations as shown above. Note that more than one operation is in some of the problems. When simplifying, use the order of operations (do what is in parentheses first) and show each step.
a. 3 λ (5 λ 2) b. 3 Θ (5 Θ 2) c. 3 [latex]\otimes[/latex] (4 [latex]\otimes[/latex] 2)
d. 8 [latex]\dagger[/latex] (9 [latex]\dagger[/latex] 6) e. (8 [latex]\dagger[/latex] 9) [latex]\dagger[/latex] 6 f. (4 ◊ 3) ◊ 2
g. 2 @ (3 [latex]\dagger[/latex] 4) h. (2 @ 3) [latex]\dagger[/latex] 4 i. (1 Θ 5) [latex]\dagger[/latex] 4
Solutions:
a. 3 λ (5 λ 2)
Solution
3 λ [ 3( 5 )( 2) ]
3 λ 30
3( 3 )( 30)
270
b. 3 Θ (5 Θ 2)
Solution
3 Θ [ 2( 5 ) + 2( 2 )]
3 Θ [ 10 + 4 ]
3 Θ 14
2( 3 ) + 2 ( 14 )
6 + 28
34
c. 3 [latex]\otimes[/latex] (4 [latex]\otimes[/latex] 2)
Solution
3 [latex]\otimes[/latex] ( 42 + 2 )
3 [latex]\otimes[/latex] (16 + 2 )
3 [latex]\otimes[/latex] 18
32 + 18
9 + 18
27
d. 8 [latex]\dagger[/latex] (9 [latex]\dagger[/latex] 6)
Solution
8 [latex]\dagger[/latex] 6
6
e. (8 [latex]\dagger[/latex] 9) [latex]\dagger[/latex] 6
Solution
8 [latex]\dagger[/latex] 6
6
f. (4 ◊ 3) ◊ 2
Solution
( 42 + 32 ) ◊ 2
( 16 + 9 ) ◊ 2
25 ◊ 2
252 + 22
625 + 4
629
g. 2 @ (3 [latex]\dagger[/latex] 4)
Solution
2 @ 3
2 ( 3 )
6
h. (2 @ 3) [latex]\dagger[/latex] 4
Solution
[ 2( 3 ) ] [latex]\dagger[/latex] 4
6 [latex]\dagger[/latex] 4
4
i. (1 Θ 5) [latex]\dagger[/latex] 4
Solution
[ 2( 1 ) + 2( 5 ) ] [latex]\dagger[/latex] 4
[ 2 + 10 ] [latex]\dagger[/latex] 4
12 [latex]\dagger[/latex] 4
4
The Associative Property
Key Takeaway – The Associative Property
An operation, [latex]\bowtie[/latex], is associative if (X [latex]\bowtie[/latex] Y) [latex]\bowtie[/latex] Z = X [latex]\bowtie[/latex] (Y [latex]\bowtie[/latex] Z) for values of X, Y, and Z.
Again, [latex]\bowtie[/latex] is just a “dummy” operation and “X” and “Y” and “Z” are dummy variables. For a particular operation to be associative, the equation must always be true no matter what values are used for X, Y, and Z.
For example, the operation [latex]*[/latex] is associative only if [latex](v * w) * x = v * (w * x)[/latex] is always true no matter what values are put in for v, w, or x.
To show that an operation is not associative, all you need to do is provide a counterexample (using actual numbers) that shows the equation is not true for at least those particular numbers. To prove an operation is associative is more involved because you must prove it is always true no matter what values you use. You would have to switch the parentheses, and show algebraically that both expressions always simplify to the same thing.
Example 5
Let [latex]@[/latex] be defined as follows: [latex]m @ n = 2n[/latex]. Is [latex]@[/latex] associative?
Solution:
If @ is associative, then (a @ b) @ c = a @ (b @ c) for all values a, b, and c. First, I’d try some numbers in for a, b, and c to see if I might come up with a counterexample:
(2 @ 3) @ 4 = [ 2(3) ] @ 4 = 6 @ 4 = 2( 4 ) = 8
2 @ (3 @ 4) = 2 @ [ 2( 4 ) ] = 2 @ 8 = 2( 8 ) = 16
This shows @ is not associative and provides us with a counterexample: Since (2 @ 3) @ 4[latex]\neq[/latex] 2 @ (3 @ 4), then @ is not associative.
Example 6
We’ll use the operation from example 3 again. Let [latex]\Omega[/latex] be defined as follows: [latex]m \Omega n = 2mn[/latex]. Is [latex]\Omega[/latex] associative?
Solution:
If [latex]\Omega[/latex] is associative, then (a [latex]\Omega[/latex] b) [latex]\Omega[/latex] c = a [latex]\Omega[/latex] (b [latex]\Omega[/latex] c) for all values a, b and c. First, I’d try some numbers in for a, b and c to see if I might come up with a counterexample:
(2 [latex]\Omega[/latex] 3) [latex]\Omega[/latex] 4 = [ 2( 2 )( 3 ) ] [latex]\Omega[/latex] 4 = 12 [latex]\Omega[/latex] 4 = 2( 12 )( 4) = 96
2 [latex]\Omega[/latex] (3 [latex]\Omega[/latex] 4) = 2 [latex]\Omega[/latex] [ 2( 3 )( 4 )] = 2 [latex]\Omega[/latex] 24 = 2( 2 )( 24 ) = 96
No counterexample here.
I might want to try another example with numbers, or I can go directly to using algebra to see if I can prove that it is always true that (a [latex]\Omega[/latex] b) [latex]\Omega[/latex] c = a [latex]\Omega[/latex] (b [latex]\Omega[/latex] c).
First, we need to simplify the left side: (a [latex]\Omega[/latex] b) [latex]\Omega[/latex] c = 2ab [latex]\Omega[/latex] c = 4abc.
Now, we have to simplify the right side: a [latex]\Omega[/latex](b [latex]\Omega[/latex] c)=a [latex]\Omega[/latex] 2bc = 4abc.
Since (a [latex]\Omega[/latex] b) [latex]\Omega[/latex] c = a [latex]\Omega[/latex] (b [latex]\Omega[/latex] c), then [latex]\Omega[/latex] is associative.
Exercise 5
For each operation listed, determine whether it is associative or not. If it is not associative, give a counterexample, like I did for [latex]@[/latex] and [latex]\Omega[/latex] If it is associative, prove it is associative, like I did for [latex]\Omega[/latex] above. Begin each problem by stating the general equation that is true if the operation listed is associative. (Answers may vary with different variables used.)
a. [latex]∇[/latex] is defined like this: [latex]m ∇ n = 2[/latex].
Write the general equation that is true if [latex]∇[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
p ∇ (m ∇ n) = (p ∇ m) ∇ n
If you answered yes, prove [latex]∇[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 ∇ ( 4 ∇ 5 ) = ( 3 ∇ 4 ) ∇ 5
3 ∇ 2 = 2 ∇ 5
2 = 2
b. [latex]λ[/latex] is defined like this: [latex]m λ n = 3mn[/latex].
Write the general equation that is true if [latex]λ[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
p λ (m λ n) = (p λ m) λ n
If you answered yes, prove [latex]λ[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 λ ( 4 λ 5 ) = ( 3 λ 4 ) λ 5
3 λ [ 3( 4 )( 5 ) ] = [ 3( 3 )( 4 )] λ 5
3 λ 60 = 36 λ 5
3( 3 )( 60 ) = 3( 36 )( 5 )
540 = 540
c. [latex]\dagger[/latex] is defined like this: [latex]m \dagger n[/latex] = the smaller value of m or n.
Write the general equation that is true if [latex]\dagger[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
p [latex]\dagger[/latex] (m [latex]\dagger[/latex] n) = (p [latex]\dagger[/latex] m) [latex]\dagger[/latex] n
If you answered yes, provide an example. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 [latex]\dagger[/latex] ( 4 [latex]\dagger[/latex] 5 ) = ( 3 [latex]\dagger[/latex] 4 ) [latex]\dagger[/latex] 5
3 [latex]\dagger[/latex] 4 = 3 [latex]\dagger[/latex] 5
3 = 3
d. [latex]Θ[/latex] is defined like this: [latex]m Θ n = 2m + 2n[/latex].
Write the general equation that is true if [latex]Θ[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
Θ is not associative
If you answered yes, prove [latex]Θ[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 Θ ( 4 Θ 5) ≠ ( 3 Θ 4 ) Θ 5
3 Θ [ 2( 4 ) + 2( 5 ) ] ≠ [ 2( 3 ) + 2 ( 4 ) ] Θ 5
3 Θ [ 8 + 10 ] ≠ [ 6 + 8 ] Θ 5
3 Θ 18 ≠ 14 Θ 5
2 ( 3 ) + 2 ( 18 ) ≠ 2 ( 14 ) + 2 ( 5 )
6 + 36 ≠ 28 + 10
42 ≠ 38
e. [latex]\otimes[/latex] is defined like this: [latex]m \otimes n = m^2 + n[/latex].
Write the general equation that is true if [latex]\otimes[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
[latex]\otimes[/latex] is not associative
If you answered yes, prove [latex]\otimes[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 [latex]\otimes[/latex] (4 [latex]\otimes[/latex] 5 ) ≠ (3 [latex]\otimes[/latex] 4 ) [latex]\otimes[/latex] 5
3 [latex]\otimes[/latex] ( 42 + 5 ) ≠ ( 32 + 4 ) [latex]\otimes[/latex] 5
3 [latex]\otimes[/latex] ( 16 + 5 ) ≠ ( 9 + 4 ) [latex]\otimes[/latex] 5
3 [latex]\otimes[/latex] 21 ≠ 13 [latex]\otimes[/latex] 5
32 + 21 ≠ 132 + 5
9 + 21 ≠ 169 + 5
30 ≠ 174
f. [latex]◊[/latex] is defined like this: [latex]m ◊ n = m^2 + n^2[/latex].
Write the general equation that is true if [latex]◊[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
◊ is not associative
If you answered yes, prove [latex]◊[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 ◊ ( 4 ◊ 5 ) ≠ ( 3 ◊ 4 ) ◊ 5
3 ◊ ( 42 + 52 ) ≠ ( 32 + 42 ) ◊ 5
3 ◊ ( 16 + 25 ) ≠ ( 9 + 16 ) ◊ 5
3 ◊ 41 ≠ 25 ◊ 5
32 + 412 ≠ 252 + 52
9 + 1681 ≠ 625 + 25
1690 ≠ 650
g. [latex]\perp[/latex] is defined like this: [latex]m \perp n = 3m + 2n + 8[/latex]
Write the general equation that is true if [latex]\perp[/latex] is associative: [latex]\underline{\qquad\qquad}[/latex]
Solution
[latex]\perp[/latex] is not associative
If you answered yes, prove [latex]\perp[/latex] is associative. If you answered no, provide a counterexample to illustrate it is not associative.
Solution
3 [latex]\perp[/latex] ( 4 [latex]\perp[/latex] 5 ) ≠ ( 3 [latex]\perp[/latex] 4 ) [latex]\perp[/latex] 5
3 [latex]\perp[/latex] [ 3( 4 ) + 2( 5 ) + 8 ] ≠ [ 3( 3 ) + 2( 4 ) + 8 ] [latex]\perp[/latex] 5
3 [latex]\perp[/latex] [ 12 + 10 + 8 ] ≠ [ 9 + 8 + 8 ] [latex]\perp[/latex] 5
3 [latex]\perp[/latex] 30 ≠ 25 [latex]\perp[/latex] 5
3( 3 ) + 2( 30 ) + 8 ≠ 3( 25 ) + 2( 5 ) + 8
9 + 60 + 8 ≠ 75 + 10 + 8
77 ≠ 93
The Distributive Property
Key Takeaway – The Distributive Property of Addition
An operation, [latex]\bowtie[/latex], distributes over another operation, [latex]\phi[/latex] if for any values of X, Y, and Z:
X [latex]\bowtie[/latex] (Y [latex]\phi[/latex] Z) = (X [latex]\bowtie[/latex] Y) [latex]\phi[/latex] (X [latex]\bowtie[/latex] Z). This is a Left-Hand Distributive Property, because the symbol on the LEFT (in this case an X) is being distributed across the parentheses to the right.
The Right-Hand Distributive Property states: An operation, [latex]\bowtie[/latex], distributes over another operation, [latex]\phi[/latex] if for any values of X, Y and Z: (Y [latex]\phi[/latex] Z) [latex]\bowtie[/latex] X = (Y [latex]\bowtie[/latex] X) [latex]\phi[/latex] (Z [latex]\bowtie[/latex] X).
Unless otherwise stated, assume that the distributive property refers to the Left-Hand Distributive Property.
Remember that [latex]\bowtie[/latex] and [latex]\phi[/latex] are just “dummy” operations and “X” and “Y” and “Z” are dummy variables. For a particular operation to be distributive over another operation, the equation must always be true no matter what values or variables are used for X, Y, and Z.
For example, the operation [latex]*[/latex] distributes over [latex]+[/latex] only if [latex]v * (w + x) = (v * w) + (v * x)[/latex] is always true no matter what value you put in for v, w, and x.
To show that an operation does not distribute over another operation, you only need to provide a counterexample (using actual numbers) that shows the equation is not true for at least those particular numbers. To prove an operation does distribute over another operation is more involved because you must prove it is always true no matter what values you use. You would first have to work the left side of the equation (by using order of operations — simplifying in parentheses first), and then work the right side of the equation (by using order of operations to simplify in parentheses first), and finally you would need to show algebraically that both expressions always simplify to the same thing.
Example 7
Let [latex]@[/latex] be defined as follows: [latex]m @ n = 2n[/latex]. We’re going to determine if [latex]@[/latex] distributes over addition.
Solution:
Write the equation that would be true if @ distributed over addition:
If @ distributes over addition, then a @ (b + c) = (a @ b) + (a @ c) for all values a, b and c. First, I’d try some numbers in for a, b and c to see if I might come up with a counterexample:
5 @ (3 + 4) = 5 @ 7 = 2( 7 ) = 14 and (5 @ 3) + (5 @ 4) = 2( 3 ) + 2( 4 ) = 6 + 8 = 14. No counterexample here.
I can try another example with numbers or try proving it algebraically.
First, simplify the left side using the definition of @: a @ (b + c) = 2(b + c) = 2b + 2c.
Now to simplify the right side: (a @ b) + (a @ c) = 2b + 2c.
Since both expressions equal the same thing (2a + 2b), a @ (b + c) = (a @ b) + (a @ c), and therefore, we say that YES, @ distributes over addition.
Example 8
Let [latex]@[/latex] be defined as follows: [latex]m @ n = 2n[/latex]. We are going to determine if addition distributes over [latex]@[/latex].
Solution:
Write the equation that is true if addition distributes over @:
If addition distributes over @, then a + (b @ c) = (a + b) @ (a + c) for all values a, b and c.
First, I’d try some numbers in for a, b and c to see if I might come up with a counterexample:
5 + (3 @ 4) = 5 + 2( 4 ) = 5 + 8 = 13 and (5 + 3) @ (5 + 4) = 8 @ 9 = 2( 9 ) = 18.
This shows that addition does not distribute over @ and provides us with a counterexample.
Since 5 + (3 @ 4)[latex]\neq[/latex] (5 + 3) @ (5 + 4), addition does not distribute over @.
Example 9
Let’s try another problem with different operations. Let [latex]\Omega[/latex] be defined by: [latex]m \Omega n = 2mn[/latex] and let [latex]$[/latex] be defined by: [latex]m $ n = m^2[/latex]. We are going to determine if [latex]\Omega[/latex] distributes over [latex]$[/latex] or if [latex]$[/latex] distributes over [latex]\Omega[/latex].
Solution:
First, let’s determine if [latex]\Omega[/latex] distributes over [latex]$[/latex].
a. Write the equation that would be true if [latex]\Omega[/latex] distributed over [latex]$[/latex]:
If [latex]\Omega[/latex] distributes over [latex]$[/latex], then this equation is true: a [latex]\Omega[/latex] (b [latex]$[/latex] c) = (a [latex]\Omega[/latex] b) [latex]$[/latex] (a [latex]\Omega[/latex] c).
Let’s compute each side of the equation by putting in some values for a, b and c to see if we find a counterexample. We’ll see if 2 [latex]\Omega[/latex] (3 [latex]$[/latex] 4) and (2 [latex]\Omega[/latex] 3) [latex]$[/latex] (2 [latex]\Omega[/latex] 4) are equal.
2 [latex]\Omega[/latex] (3 [latex]$[/latex] 4) = 2 [latex]\Omega[/latex] 32 = 2 [latex]\Omega[/latex] 9 = 2( 2 )( 9 ) = 36
(2 [latex]\Omega[/latex] 3) [latex]$[/latex] (2 [latex]\Omega[/latex] 4) = [ 2( 2 )( 3 ) ] [latex]$[/latex] [ 2( 2 )( 4 ) ] = 12 [latex]$[/latex] 16 = 122 = 144
We have a counterexample. Therefore, [latex]\Omega[/latex] does not distribute over [latex]$[/latex].
Next, we’ll determine if [latex]$[/latex] distributes over [latex]\Omega[/latex].
b. Write the equation that is true if [latex]$[/latex] distributes over [latex]\Omega[/latex]:
If [latex]$[/latex] distributes over [latex]\Omega[/latex], then this equation is true for all values of a, b and c: a [latex]$[/latex] (b [latex]\Omega[/latex] c) = (a [latex]$[/latex] b) [latex]\Omega[/latex] (a [latex]$[/latex] c).
Let’s compute each side of the equation by putting in some values for a, b and c to see if we find a counterexample. Let’s see if 2 [latex]$[/latex] (3 [latex]\Omega[/latex] 4) and (2 [latex]$[/latex] 3) [latex]\Omega[/latex] (2 [latex]$[/latex] 4) are equal.
2 [latex]$[/latex] (3 [latex]\Omega[/latex] 4) = 2 [latex]$[/latex] [ 2( 3 )( 4 ) ] = 2 [latex]$[/latex] 24 = 22 = 4
(2 [latex]$[/latex] 3) [latex]\Omega[/latex] (2 [latex]$[/latex] 4) = 22 [latex]\Omega[/latex] 22 = 4 [latex]\Omega[/latex] 4 = 2( 4 )( 4) = 32
We have a counterexample. Therefore, [latex]$[/latex] does not distribute over [latex]\Omega[/latex].
Exercise 6
Let [latex]∇[/latex] and [latex]λ[/latex] be defined as follows: [latex]a ∇ b = 2[/latex] and [latex]a λ b = 3ab[/latex].
a. Write the general equation that is true if ∇ distributes over λ.
Solution
∇ does not distribute over λ
If ∇ distributes over λ, prove it. Otherwise, provide a counterexample to illustrate that !does not distribute over λ.
Solution
3 ∇ ( 4 λ 5 ) ≠ ( 3 ∇ 4 ) λ ( 3 ∇ 5 )
3 ∇ [ 3( 4 )( 5) ] ≠ 2 λ 2
3 ∇ 60 ≠ 3( 2 )( 2 )
2 ≠ 12
b. Write the general equation that is true if λ distributes over ∇.
Solution
λ does not distribute over ∇
If λ distributes over ∇, prove it. Otherwise, provide a counterexample to illustrate that λ does not distribute over ∇.
Solution
3 λ ( 4 ∇ 5 ) ≠ ( 3 λ 4 ) ∇ ( 3 λ 5 )
3 λ ( 2 ) ≠ [ 3( 3 )( 4 ) ] ∇ [ 3( 3 )( 5 )]
3( 3 )( 2 ) ≠ 36 ∇ 45
18 ≠ 2
Exercise 7
Write the general equation that is true if addition distributes over multiplication.
Solution
Addition does not distribute over multiplication.
If addition distributes over multiplication, prove it. Otherwise, provide a counterexample to illustrate that addition does not distribute over multiplication.
Solution
2 + ( 3 × 4 ) ≠ ( 2 + 3 ) × ( 2 + 4 )
2 + 12 ≠ 5 × 6
14 ≠ 30
Exercise 8
Write the general equation that is true if multiplication distributes subtraction.
Solution
a × (b – c) = a × b – a × c
If multiplication distributes over subtraction, prove it. Otherwise, provide a counterexample to illustrate that multiplication does not distribute over subtraction.
Solution
2 × ( 4 – 3 ) = ( 2 × 4 ) – ( 2 × 3 )
2 × 1 = 8 – 6
2 = 2
Exercise 9
Let [latex]◊[/latex] and [latex]@[/latex] be defined as follows: [latex]a ◊ b = a^2 + b^2[/latex] and [latex]a @ b = 2b[/latex]
Write the general equation that is true if ◊ distributes over @.
Solution
◊ does not distribute over @
If ◊ distributes over @, prove it. Otherwise, provide a counterexample to illustrate that ◊ does not distribute over @.
Solution
2 ◊ ( 3 @ 4 ) ≠ ( 2 ◊ 3 ) @ ( 2 ◊ 4 )
2 ◊ 2( 4 ) ≠ ( 22 + 33 ) @ ( 22 + 42 )
2 ◊ 8 ≠ ( 4 + 9 ) @ ( 4 + 16 )
22 + 82 ≠ 13 @ 20
4 + 64 ≠ 2( 20 )
68 ≠ 40
Write the general equation that is true if @ distributes over ◊.
Solution
@ does not distribute over ◊
If @ distributes over ◊, prove it. Otherwise, provide a counterexample to illustrate that @ does not distribute over ◊.
Solution
2 @ ( 3 ◊ 4 ) ≠ ( 2 @ 3 ) ◊ ( 2 @ 4 )
2 @ ( 32 + 42 ) ≠ 2( 3 ) ◊ 2( 4 )
2 @ ( 9 + 16 ) ≠ 6 ◊ 8
2 @ 25 ≠ 62 + 82
2( 25 ) ≠ 36 + 64
50 ≠ 100
Exercise 10
Make up and define two new operations.
Write a general equation that is true if one operation distributes over the other one.
Determine if the distributive property holds for your operations by proving it or providing a counterexample illustrating the equation in part b is not true.
Exercise 11
Define [latex]\phi[/latex] and [latex]\wedge[/latex] as follows: [latex]m \phi n = 2m + 3n[/latex] and [latex]m \wedge n = mn + 2[/latex]
a.
State the equation that is true if [latex]\phi[/latex] is commutative.
Solution
[latex]\phi[/latex] is not commutative
Prove it is commutative or provide a counterexample if it is not commutative.
Solution
2 [latex]\phi[/latex] 3 ≠ 3 [latex]\phi[/latex] 2
2( 2 ) + 3( 3 ) ≠ 2( 3 ) + 3( 2 )
4 + 9 ≠ 6 + 6
13 ≠ 12
b.
State the equation that is true if [latex]\wedge[/latex] is commutative.
Solution
m [latex]\wedge[/latex] n = n [latex]\wedge[/latex] m
Prove it is commutative or provide a counterexample if it is not commutative.
Solution
2 [latex]\wedge[/latex] 3 = 3 [latex]\wedge[/latex] 2
2 × 3 + 2 = 3 × 2 + 2
6 + 2 = 6 + 2
8 = 8
c.
State the equation that is true if [latex]\phi[/latex] is associative.
Solution
[latex]\phi[/latex] is not associative
Prove it is associative or provide a counterexample if it is not associative.
Solution
2 [latex]\phi[/latex] ( 3 [latex]\phi[/latex] 4 ) ≠ ( 2 [latex]\phi[/latex] 3 ) [latex]\phi[/latex] 4
2 [latex]\phi[/latex] [ 2( 3 ) + 3( 4 ) ] ≠ [ 2( 2 ) + 3( 3 ) ] [latex]\phi[/latex] 4
2 [latex]\phi[/latex] [ 6 + 12 ] ≠ [ 4 + 9 ] [latex]\phi[/latex] 4
2 [latex]\phi[/latex] 18 ≠ 13 [latex]\phi[/latex] 4
2( 2 ) + 3( 18 ) ≠ 2( 13 ) + 3( 4 )
4 + 54 ≠ 26 + 12
58 ≠ 38
d.
State the equation that is true if [latex]\wedge[/latex] is associative.
Solution
[latex]\wedge[/latex] is not associative
Prove it is associative or provide a counterexample if it is not associative.
Solution
2 [latex]\wedge[/latex] (3 [latex]\wedge[/latex] 4 ) ≠ ( 2 [latex]\wedge[/latex] 3 ) [latex]\wedge[/latex] 4
2 [latex]\wedge[/latex] ( 3 × 4 + 2 ) ≠ ( 2 × 3 + 2 ) [latex]\wedge[/latex] 4
2 [latex]\wedge[/latex] ( 12 + 2 ) ≠ ( 6 + 2 ) [latex]\wedge[/latex] 4
2 [latex]\wedge[/latex] 14 ≠ 8 [latex]\wedge[/latex] 4
2 × 14 + 2 ≠ 8 × 4 + 2
28 + 2 ≠ 32 + 2
30 ≠ 34
e.
State the equation that is true if [latex]\phi[/latex] distributes over addition.
Solution
[latex]\phi[/latex] does not distribute over addition
Prove it is distributive over addition or provide a counterexample if it is not distributive over addition.
Solution
2 [latex]\phi[/latex] ( 3 + 4 ) ≠ ( 2 [latex]\phi[/latex] 3 ) + ( 2 [latex]\phi[/latex] 4 )
2 [latex]\phi[/latex] 7 ≠ [ 2( 2 ) + 3( 3 ) ] + [ 2( 2 ) + 3( 4 ) ]
2( 2 ) + 3( 7 ) ≠ [ 4 + 9 ] + [ 4 + 12 ]
4 + 21 ≠ 13 + 16
25 ≠ 29
f.
State the equation that is true if [latex]\wedge[/latex] distributes over addition.
Solution
[latex]\wedge[/latex] does not distribute over addition
Prove it is distributive over addition or provide a counterexample if it is not distributive over addition..
Solution
2 [latex]\wedge[/latex] ( 3 + 4 ) ≠ ( 2 [latex]\wedge[/latex] 3 ) + ( 2 [latex]\wedge[/latex] 4 )
2 [latex]\wedge[/latex] 7 ≠ ( 2 × 3 + 2 ) + ( 2 × 4 + 2 )
2 × 7 + 2 ≠ ( 6 + 2 ) + ( 8 + 2 )
14 + 2 ≠ 8 + 10
16 ≠ 18
g.
State the equation that is true if [latex]\phi[/latex] distributes over [latex]\wedge[/latex].
Solution
[latex]\phi[/latex] does not distribute over [latex]\wedge[/latex]
Prove [latex]\phi[/latex] distributes over [latex]\wedge[/latex] or provide a counterexample if [latex]\phi[/latex] does not distribute over [latex]\wedge[/latex].
Solution
2 [latex]\phi[/latex] ( 3 [latex]\wedge[/latex] 4 ) ≠ ( 2 [latex]\phi[/latex] 3 ) [latex]\wedge[/latex] ( 2 [latex]\phi[/latex] 4 )
2 [latex]\phi[/latex] ( 3 × 4 + 2 ) ≠ [ 2( 2 ) + 3( 3 ) ] [latex]\wedge[/latex] [ 2( 2 ) + 3( 4 ) ]
2 [latex]\phi[/latex] ( 12 + 2 ) ≠ [ 4 + 9 ] [latex]\wedge[/latex] [ 4 + 12 ]
2 [latex]\phi[/latex] 14 ≠ 13 [latex]\wedge[/latex] 16
2( 2 ) + 3( 14 ) ≠ 13 × 16 + 2
4 + 42 ≠ 208 + 2
46 ≠ 210
h.
State the equation that is true if [latex]\wedge[/latex] distributes over [latex]\phi[/latex].
Solution
[latex]\wedge[/latex] does not distribute over [latex]\phi[/latex]
Prove [latex]\wedge[/latex] distributes over [latex]\phi[/latex] or provide a counterexample if [latex]\wedge[/latex] does not distribute over [latex]\phi[/latex].
Solution
2 [latex]\wedge[/latex] ( 3 [latex]\phi[/latex] 4 ) ≠ ( 2 [latex]\wedge[/latex] 3 ) [latex]\phi[/latex] ( 2 [latex]\wedge[/latex] 4 )
2 [latex]\wedge[/latex] [ 2( 3 ) + 3( 4 ) ] ≠ ( 2 × 3 + 2 ) [latex]\phi[/latex] ( 2 × 4 + 2 )
2 [latex]\wedge[/latex] [ 6 + 12 ] ≠ ( 6 + 2 ) [latex]\phi[/latex] ( 8 + 2 )
2 [latex]\wedge[/latex] 18 ≠ 8 [latex]\phi[/latex] 10
2 × 18 + 2 ≠ 2( 8 ) + 3( 10 )
36 + 2 ≠ 16 + 30
38 ≠ 46
For these last few exercises, you’ll be working with the Right-Hand Distributive Property. For clarification, “[latex]\perp[/latex] right-hand distributes over [latex]@[/latex]” means the same thing as “[latex]\perp[/latex] distributes over [latex]@[/latex], using the Right-Hand Distributive Property.”
Again, the Right-Hand Distributive Property is: An operation, [latex]\bowtie[/latex], distributes over another operation, [latex]\phi[/latex] if for any values of X, Y, and Z: (Y [latex]\phi[/latex] Z) [latex]\bowtie[/latex] X = (Y [latex]\bowtie[/latex] X) [latex]\phi[/latex] (Z [latex]\bowtie[/latex] X).
Exercise 12
State the equation that is true if multiplication right-hand distributes over addition.
Solution
(a + b) × c = (a × c) + (b × c)
If multiplication right-hand distributes over addition, provide an example. Otherwise, provide a counterexample if multiplication does not right-hand distribute over addition.
Solution
( 2 + 3 ) × 4 = ( 2 × 4 ) + ( 3 × 4 )
5 × 4 = 8 + 12
20 = 20
Exercise 13
State the equation that is true if addition right-hand distributes over multiplication.
Solution
Addition right-hand does not distribute over multiplication.
Prove addition right-hand distributes over multiplication or provide a counterexample if addition does not right-hand distribute over multiplication.
Solution
( 2 × 3 ) + 4 ≠ ( 2 + 4 ) × ( 3 + 4 )
6 + 4 ≠ 6 × 7
10 ≠ 42
Exercise 14
State the equation that is true if division right-hand distributes over addition.
Solution
(a + b) ÷ c = (a ÷ c) + (b ÷ c)
If division right-hand distributes over addition, provide an example. Otherwise, provide a counterexample if division does not right-hand distribute over addition.
Solution
( 8 + 10 ) ÷ 2 = ( 8 ÷ 2 ) + ( 10 ÷ 2 )
18 ÷ 2 = 4 + 5
9 = 9
Exercise 15
State the equation that is true if division left-hand distributes over addition.
Solution
Division left-hand does not distribute over addition.
Prove division left-hand distributes over addition or provide a counterexample if division does not left-hand distribute over addition.
Solution
30 ÷ ( 2 + 3 ) ≠ ( 30 ÷ 2 ) + ( 30 ÷ 3 )
30 ÷ 5 ≠ 15 + 10
6 ≠ 25
Exercise 16
Define [latex]\phi[/latex] and [latex]\wedge[/latex] as follows: [latex]m \phi n = 2m + 3n[/latex] and [latex]m \wedge n = mn + 2[/latex]
a. State the equation that is true if [latex]\phi[/latex] right-hand distributes over addition.
Solution
[latex]\phi[/latex] does not right-hand distribute over addition.
Prove [latex]\phi[/latex] right-hand distributes over addition or provide a counterexample if [latex]\phi[/latex] does not right-hand distribute over addition.
Solution
( 3 + 4 ) [latex]\phi[/latex] 2 ≠ (3 [latex]\phi[/latex] 2 ) + ( 4 [latex]\phi[/latex] 2 )
7 [latex]\phi[/latex] 2 ≠ [ 2( 3 ) + 3( 2 ) ] + [ 2( 4 ) + 3( 2 ) ]
2( 7 ) + 3( 2 ) ≠ [ 6 + 6 ] + [ 8 + 6 ]
14 + 6 ≠ 12 + 14
20 ≠ 26
b. State the equation that is true if [latex]\wedge[/latex] right-hand distributes over addition.
Solution
[latex]\wedge[/latex] does not right-hand distribute over addition.
Prove [latex]\wedge[/latex] right-hand distributes over addition or provide a counterexample if [latex]\wedge[/latex] does not right-hand distribute over addition.
Solution
( 3 + 4 ) [latex]\wedge[/latex] 2 ≠ (3 [latex]\wedge[/latex] 2 ) + ( 4 [latex]\wedge[/latex] 2 )
7 [latex]\wedge[/latex] 2 ≠ ( 3 × 2 + 2 ) + ( 4 × 2 + 2 )
7 × 2 + 2 ≠ 8 + 10
16 ≠ 18
c. State the equation that is true if [latex]\phi[/latex] right-hand distributes over [latex]\wedge[/latex].
Solution
[latex]\phi[/latex] does not right-hand distribute over [latex]\wedge[/latex].
Prove [latex]\phi[/latex] right-hand distributes over [latex]\wedge[/latex] or provide a counterexample if [latex]\phi[/latex] does not right-hand distribute over [latex]\wedge[/latex].
Solution
( 3 [latex]\wedge[/latex] 4 ) [latex]\phi[/latex] 2 ≠ (3 [latex]\phi[/latex] 2 ) [latex]\wedge[/latex] ( 4 [latex]\phi[/latex] 2 )
( 3 × 4 + 2 ) [latex]\phi[/latex] 2 ≠ [ 2( 3) + 3( 2 ) ] [latex]\wedge[/latex] [ 2( 4 ) + 3( 2 ) ]
14 [latex]\phi[/latex] 2 ≠ [ 6 + 6 ] [latex]\wedge[/latex] [ 8 + 6 ]
2( 14 ) + 3( 2 ) ≠ 12 [latex]\wedge[/latex] 14
28 + 6 ≠ 12 × 14 + 2
34 ≠ 170
d. State the equation that is true if [latex]\wedge[/latex] right-hand distributes over [latex]\phi[/latex].
Solution
[latex]\wedge[/latex] does not right-hand distribute over [latex]\phi[/latex].
Prove [latex]\wedge[/latex] right-hand distributes over [latex]\phi[/latex] or provide a counterexample if [latex]\wedge[/latex] does not right-hand distribute over [latex]\phi[/latex].
Solution
( 3 [latex]\phi[/latex] 4 ) [latex]\wedge[/latex] 2 ≠ (3 [latex]\wedge[/latex] 2 ) [latex]\phi[/latex] ( 4 [latex]\wedge[/latex] 2 )
[ 2( 3 ) + 3( 4 ) ] [latex]\wedge[/latex] 2 ≠ ( 3 × 2 + 2 ) [latex]\phi[/latex] ( 4 × 2 + 2 )
[ 6 + 12 ] [latex]\wedge[/latex] 2 ≠ 8 [latex]\phi[/latex] 10
18 [latex]\wedge[/latex] 2 ≠ 2( 8 ) + 3( 10 )
18 × 2 + 2 ≠ 16 + 30
38 ≠ 46
The Closure Property
Key Takeaway – Closure Property
A set of numbers is closed under an operation when the result of that operation is also in the set.
For our examples, we’ll use the set of whole numbers as our set.
To show that an operation is not closed over the set of whole numbers, all you need to do is provide a counterexample (with particular values) that shows the equation is not true for at least those particular values. To prove an operation is closed is more involved because you must prove it is always true no matter what values you use.
To illustrate this property, let’s look at the following examples.
Suppose we use addition on the set of whole numbers. If we add any two numbers like 2 and 3, the sum of 5 is also a whole number. We can continue to add other numbers, but since the sum of any two whole numbers is a whole number, then the set is closed under addition.
Now let’s use the set {1, 2, 3, 4, 5}. If we add 1 and 2 the sum of 3 is in the set, but if we add 4 and 5 the sum is 9. Since 9 is not in the set, then the set is not closed under addition.
Exercise 17
For each operation listed, determine whether it is closed under the set of whole numbers. If it is closed, prove this closure and if it is not closed, give a counterexample.
a. m [latex]\perp[/latex] n = 3m + 2n + 8
b. m * n = m + 2n
c. m ◊ n = m2 + n2
d. m ∇ n = 2
e. m λ n = 3mn
f. m Θ n = 2m + 2n
g. m [latex]\otimes[/latex] n = m2 + n
Solutions:
a. m [latex]\perp[/latex] n = 3m + 2n + 8
Solution
[latex]\perp[/latex] is closed for all values of whole numbers m and n, since 3m and 2n are whole numbers. This means that 3m + 2n + 8 must be a whole number.
b. m * n = m + 2n
Solution
* is closed for all values of whole numbers m and n, since m and 2n are whole numbers. This means that m + 2n must be a whole number.
c. m ◊ n = m2 + n2
Solution
◊ is closed for all values of whole numbers m and n, since m2 and n2 are whole numbers. This means that m2 + n2 must be a whole number.
d. m ∇ n = 2
Solution
∇ is closed for all values of whole numbers m and n since the the solution 2 doesn’t depend on either m or n.
e. m λ n = 3mn
Solution
λ is closed for all values of whole numbers m and n. This means that 3mn must be a whole number.
f. m Θ n = 2m + 2n
Solution
Θ is closed for all values of whole numbers m and n, since 2m and 2n are whole numbers. This means that 2m + 2n must be a whole number.
g. m [latex]\otimes[/latex] n = m2 + n
Solution
[latex]\otimes[/latex] is closed for all values of whole numbers m and n, since m2 and n are whole numbers. This means that m2 + n must be a whole number.
The Identity Property
Key Takeaway – Identity Property
For an operation, [latex]\bowtie[/latex], if “E” is the identity element, then E [latex]\bowtie[/latex] X = X = X [latex]\bowtie[/latex] E for all elements X in the set.
Remember that the operation [latex]\bowtie[/latex] is just a “dummy” operation and “X” is a dummy variable. For a particular operation to have an identity element, the equation must always be true no matter what value is used for X .
For instance, the operation [latex]*[/latex] has an identity element if and only if [latex]E * n = n = n * E[/latex] is always true no matter what value is put in for m.
Recall that for addition, the identity element is 0 and for multiplication the element is 1.
Example 10
Suppose that a number system only uses the symbols A, B, C, D and the operation ♣. The basic facts are illustrated in the table.
Example: A ♣ C = C or D ♣ B = A
| ♣ | A | B | C | D |
| A | A | B | C | D |
| B | B | C | D | A |
| C | C | D | A | B |
| D | D | A | B | C |
a) Is the system closed under the operation? Why or why not?
b) Is the operation commutative? Why or why not?
c) Is there an identity element for the operation? If so, what is the identity element?
Solution:
Let’s look at the closure first. In order to be closed, the result from the operation is also in the system. As we look at the system above, the elements A, B, C, and D are the only ones to result from the operation. This means that the system is closed under the operation.
The operation is also commutative. If we look at B ♣ C we get D. We also get D from looking at C ♣ B.
The operation has an identity element and that element is A. Since A ♣ A = A, A ♣ B = B, A ♣ C = C and A ♣ D = D it proves that A is the identity element.
Exercise 18
Suppose that a number system only uses the symbols A, B, C, D and the operation Ψ. The basic facts are illustrated in the table.
Example: A Ψ C = A or D Ψ B = A
| Ψ | A | B | C | D |
| A | D | C | A | B |
| B | C | D | B | A |
| C | A | B | C | D |
| D | B | A | D | C |
a)
b) Prove why or why not.
Solution
The system uses values for A, B, C, and D. Since these are the only values in the table, the system must be closed.
c)
d) Prove why or why not.
Solution
The operation ψ is commutative. Since A ψ D = B and D ψ A = B, this shows that the operation is commutative.
e)
f) If so, what is the identity element?
Solution
The operation has an identity element and the element is C. Since A ψ C = A, B ψ C = B, C ψ C = C and D ψ C = D, this proves D is the identity element.