DG1, DG4 - Binary operations and Cayley tables

DG1, DG4 - Binary operations and Cayley tables

What changes when an operation acts on a particular set rather than on any numbers you like? You will learn what makes a rule a binary operation, calculate with modular arithmetic and matrices, and construct and read Cayley tables without reversing the operands.

Closure and table construction

Can a rule combine two allowed inputs and produce a forbidden answer? Take the set S={−1,1}S=\{-1,1\}. Adding its elements can give 22, which is outside SS. Multiplying them always gives −1-1 or 11. The set matters as much as the rule.

Binary operation and closure

A binary operation on a set SS assigns exactly one element of SS to every ordered pair (a,b)(a,b) of elements of SS. The set is closed under the rule when every such output belongs to SS.

“Binary” means two inputs, not base-two arithmetic. A familiar rule may fail on a chosen set: division is not a binary operation on the real numbers because division by zero is undefined. Removing zero makes division defined for every pair of nonzero real numbers, with a nonzero real result.

A Cayley table records every output for a finite set. We use the row heading as the first operand and the column heading as the second. Keep the printed order of the headings; do not rearrange entries by size.

Worked example 1: finding a failure of closure

Question. Does ordinary addition define a binary operation on S={−1,1}S=\{-1,1\}? Display all four outputs.

Plan. Use each element as a row input and combine it with each column input.

+−11−1−20102\begin{array}{c|rr} +&-1&1\\\hline -1&-2&0\\ 1&0&2 \end{array}

For example, row −1-1, column 11 contains −1+1=0-1+1=0. None of these outputs belongs to SS.

Answer. No: SS is not closed under addition. The displayed array records the failure; it is not a Cayley table for a binary operation on SS.

Check. The single pair (1,1)(1,1) already gives 2∉S2\notin S, independently confirming the conclusion.

Keep the set and its heading order, but change only the rule to multiplication. This time every cell must stay in the set.

Worked example 2: constructing a closed table

Question. Construct the Cayley table for ordinary multiplication on S={−1,1}S=\{-1,1\} and establish closure.

Plan. Multiply the row input by the column input in each of the four cells.

×−11−11−11−11\begin{array}{c|rr} \times&-1&1\\\hline -1&1&-1\\ 1&-1&1 \end{array}

The diagonal entries are (−1)(−1)=1(-1)(-1)=1 and 1(1)=11(1)=1; the other products are −1-1.

Answer. The table above is complete. Every output is in SS, and every ordered pair has been checked, so multiplication is a binary operation on SS.

Check. A product of two signs is positive when the signs match and negative otherwise. This reproduces all four entries without recalculating the table row by row.

A single forbidden output disproves closure; proving closure needs every pair, or a general argument covering them all.

Watch out

Checking that a rule works for one pair does not establish closure. Check all pairs in a finite table, and check that every required output is defined as well as in the set.

Try a set with one extra boundary element.

Modular arithmetic

A clock returns to its starting label after a full circuit. Modular arithmetic does the same with integer labels: after calculating, replace the result by its remainder.

For an integer n≥2n\ge2, use the representatives 0,1,…,n−10,1,\ldots,n-1. Write a≡r(modn)a\equiv r\pmod n when a−ra-r is a multiple of nn. For example, 17≡2(mod5)17\equiv2\pmod5, because 17−2=1517-2=15. This congruence is a relation between integers; an operation such as +5+_5 returns one chosen representative.

Addition and multiplication modulo n

On Sn={0,1,…,n−1}S_n=\{0,1,\ldots,n-1\},

a+nb=the remainder of a+b on division by n,a+_n b=\text{the remainder of }a+b\text{ on division by }n, a×nb=the remainder of ab on division by n.a\times_n b=\text{the remainder of }ab\text{ on division by }n.

Both outputs lie in SnS_n. These definitions and notation are not supplied in the examination formula booklet.

The subscript tells you the modulus; it does not mean multiply or add nn. A remainder is at least zero and less than nn. For a negative integer, add a suitable multiple of nn: −2≡3(mod5)-2\equiv3\pmod5, so its chosen representative is 33, not −2-2.

Worked example 3: addition modulo 5

Question. Find 4+544+_5 4, and construct the +5+_5 row headed 44, with columns 0,1,2,3,40,1,2,3,4.

Plan. Add ordinarily, then remove whole multiples of 55.

4+4=8=1(5)+3⟹4+54=3.4+4=8=1(5)+3\quad\Longrightarrow\quad4+_5 4=3.

The row is

+501234440123\begin{array}{c|rrrrr} +_5&0&1&2&3&4\\\hline 4&4&0&1&2&3 \end{array}

For example, the column headed 11 gives 4+1=54+1=5, whose remainder is 00.

Answer. 4+54=34+_5 4=3; the row is 4,0,1,2,34,0,1,2,3.

Check. Each output differs from its ordinary sum by 00 or 55, and each lies between 00 and 44.

Keep the same operands, modulus and heading order. Changing addition to multiplication changes what you calculate before reducing.

Worked example 4: multiplication modulo 5

Question. Find 4×544\times_5 4, and construct the ×5\times_5 row headed 44, with columns 0,1,2,3,40,1,2,3,4.

Plan. Multiply ordinarily, then replace each product by its remainder modulo 55.

4(4)=16=3(5)+1⟹4×54=1.4(4)=16=3(5)+1\quad\Longrightarrow\quad4\times_5 4=1.

The ordinary products are 0,4,8,12,160,4,8,12,16, giving

×501234404321\begin{array}{c|rrrrr} \times_5&0&1&2&3&4\\\hline 4&0&4&3&2&1 \end{array}

Answer. 4×54=14\times_5 4=1; the row is 0,4,3,2,10,4,3,2,1.

Check. Since 4≡−1(mod5)4\equiv-1\pmod5, multiplication by 44 reverses the sign of each column label modulo 55, giving the same row.

A smaller chosen set may fail closure even though the full residue set is closed. For instance, {1,2,3,4}\{1,2,3,4\} is not closed under +5+_5, because 1+54=01+_5 4=0 is excluded.

Watch out

Wrong: 4×54=164\times_5 4=16. The ordinary product is an intermediate result. The operation's output is the representative 11.

Apply the same two-stage process with a new modulus.

Matrix multiplication as an operation

You already know how to multiply matrices: take a row of the first matrix against a column of the second. A set of matrices can be closed under that multiplication, just as a set of integers can.

Continue this lesson.

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