Binary operations · Lesson 2 of 2

Properties and tables

Closure, commutativity and associativity, the identity element and inverses, from a rule and from a table, and building and using an operation table in modular arithmetic.

17 minYou should already know: Expressions, formulae & change of subject
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An operation can have some useful properties. Each one is tested with the operation’s rule or read from its table. A table lists the elements along the top and down the side, and the entry in row xx, column yy is x∗yx * y. The tables below use ⊗\otimes, multiplication modulo 5 on {1,2,3,4}\{1, 2, 3, 4\} (see modular arithmetic).

Closure

An operation is closed on a set if a∗ba * b is always in the set, whatever aa and bb you pick from it. In a table, every entry must be one of the elements.

⊗1234
11234
22413
33142
44321
ClosedEvery entry is 1, 2, 3 or 4: ⊗ is closed on {1, 2, 3, 4}

Commutative

An operation is commutative if a∗b=b∗aa * b = b * a for every aa and bb: the order doesn’t matter. To test a rule, swap aa and bb and see if you get the same expression. In a table, every entry must match its mirror image across the leading diagonal.

⊗1234
11234
22413
33142
44321
Commutative2 ⊗ 4 = 4 ⊗ 2 = 3: the table is symmetric about the leading diagonal

Associative

An operation is associative if (a∗b)∗c=a∗(b∗c)(a * b) * c = a * (b * c) for every aa, bb and cc: it doesn’t matter where the brackets go.

abca * b(a * b) * c=abcb * ca * (b * c)
AssociativeThe two ways of bracketing must always give the same answer

The identity element

The identity ee leaves every element unchanged: a∗e=e∗a=aa * e = e * a = a for every aa. For ordinary addition it is 0, and for multiplication it is 1.

To find it from a rule, write out a∗e=aa * e = a and solve for ee. Then check that e∗a=ae * a = a as well. In a table, the identity’s row and its column both repeat the headings.

⊗1234
11234
22413
33142
44321
IdentityRow 1 and column 1 repeat the headings: 1 is the identity

Inverses

The inverse of aa, written a−1a^{-1}, is the element that combines with aa to give the identity: a∗a−1=a−1∗a=ea * a^{-1} = a^{-1} * a = e. Find the identity first. From a rule, solve a∗x=ea * x = e. In a table, go along row aa until you find ee: its column is the inverse.

⊗1234
11234
22413
33142
44321
Inverse2 ⊗ 3 = 1, the identity, so the inverse of 2 is 3

Try it

Operation tablesPick a table and a property

The table of multiplication modulo 5 on {1, 2, 3, 4}

⊗1234
1identity
Row 1 and column 1 repeat the headings: 1 ⊗ x = x ⊗ 1 = x for every x, so 1 is the identity.

Pick a table, then a property. Look for the tables that fail: one has no identity, one is not commutative, and one is not closed.

Building and using a table

To build a table from a rule, work out each entry with the rule. If the question says modulo nn, reduce each answer to its remainder on dividing by nn. To solve an equation such as 3⊗n=23 \otimes n = 2 from the table, go along row 3, find the 2, and read off its column. On the diagonal you find n⊗nn \otimes n.

⊗1234
11234
22413
33142
44321
Solving from a table3 ⊗ n = 2: along row 3, the 2 is in column 4, so n = 4

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q13 (b)

The operation Δ\Delta is defined on the set T={2,3,5,7}T = \{2, 3, 5, 7\} by x Δ y=(x+y+xy) mod 8x\,\Delta\,y = (x + y + xy) \bmod 8. (i) Construct the modulo 8 table for the operation Δ\Delta on the set TT. (ii) Use the table to find: I. 2 Δ (5 Δ 7)2\,\Delta\,(5\,\Delta\,7); II. nn if 2 Δ n=5 Δ 72\,\Delta\,n = 5\,\Delta\,7.

  1. (i) One entry at a time

    2 Δ 3=2+3+6=112 \,\Delta\, 3 = 2 + 3 + 6 = 11, and 11=8+311 = 8 + 3, so the entry is 3. In the same way, 5 Δ 7=5+7+35=47=5×8+75 \,\Delta\, 7 = 5 + 7 + 35 = 47 = 5 \times 8 + 7, so the entry is 7.

    Think first. Work out 2 Δ 3, then reduce modulo 8.

  2. The table

    Δ\Delta2357
    20317
    33777
    51737
    77777

    Think first. Fill in all sixteen entries.

  3. (ii) I. The bracket first

    5 Δ 7=75 \,\Delta\, 7 = 7, so 2 Δ (5 Δ 7)=2 Δ 7=72 \,\Delta\, (5 \,\Delta\, 7) = 2 \,\Delta\, 7 = 7.

    Think first. Read 5 Δ 7 from the table, then use it.

  4. II. Read the row

    In row 2 the entries are 0, 3, 1, 7. The 7 is in the column for 7, so n=7n = 7.

    Think first. 2 Δ n = 7: where is 7 in row 2?

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q13 (b)

An operation (∗)(*) is defined on the set R\mathbb R of real numbers by m∗n=−nm2+1m * n = \dfrac{-n}{m^2 + 1}, where m,n∈Rm, n \in \mathbb R. If m=−3m = -3 and n=−10n = -10, show whether or not (∗)(*) is commutative.

  1. Work out m * n

    m∗n=−(−10)(−3)2+1=1010=1m * n = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1.

    Think first. Put m = −3 and n = −10 into the rule.

  2. Swap them

    n∗m=−(−3)(−10)2+1=3101n * m = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}.

    Think first. Now the first number is −10 and the second is −3.

  3. Compare

    1≠31011 \ne \frac{3}{101}, so ∗* is not commutative.

    Think first. Are they equal?

Your turn

WAEC 2015 · Paper 2 · Q11 (c)

  1. (c)

    An operation ⊕\oplus is defined on the set X={1,3,5,6}X = \{1, 3, 5, 6\} by m⊕n=m+n+2(mod7)m \oplus n = m + n + 2 \pmod 7, where m,n∈Xm, n \in X. (i) Draw a table for the operation. (ii) Using the table, find the truth set of: I. 3⊕n=33 \oplus n = 3; II. n⊕n=3n \oplus n = 3.

    Model answer

    (i) Work out m+n+2m + n + 2 and take the remainder on dividing by 7:

    ⊕\oplus 1 3 5 6
    1 4 6 1 2
    3 6 1 3 4
    5 1 3 5 6
    6 2 4 6 0

    (ii) I. In the row for 3, the entry 3 is under n=5n = 5: truth set {5}\{5\}. II. The diagonal (n⊕nn \oplus n) reads 4,1,5,04, 1, 5, 0 and never 3: truth set {}\{\} (empty).

Worked solution (try it first)

(c)(i)

  1. Work out m+n+2m + n + 2, then take the remainder when dividing by 7.
  2. For example, 5⊕6=13=7+65 \oplus 6 = 13 = 7 + 6, so 5⊕6=65 \oplus 6 = 6.
  3. ⊕\oplus 1 3 5 6
    1 4 6 1 2
    3 6 1 3 4
    5 1 3 5 6
    6 2 4 6 0

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