The next result is a game changer! It tells us that there is a unique element (a^{-1}) such that (aa^{-1} = _{b} 1) if and only if (a) is in the reduced set of residues (modulo (b)). Thus division is well-defined in the reduced set of residues modulo (b). A ring is a structure with addition and its inverse subtraction plus multiplication, but where multiplication may not have an inverse. A more detailed description of these algebraic construction is given in Section **??**. The numbers (1) and (-1) are always in the reduced set of residues modulo (b). This set is sometimes called the set of units (see Definition **??**) of (mathbb{Z}_{b}).

Proposition 5.16

Let (mathbb{R}) be a reduced set of residues modulo (b). Then

- for every (a in mathbb{R}), there is a unique (a') in (mathbb{R}) such that (a′a = _{b} aa′ = _{b} 1)
- for every (a otin mathbb{R}), there exists no (x in mathbb{Z}_{b}) such that (ax = _{b} 1)
- let (mathbb{R} = {x_{i}}_{i=1}^{varphi (b)}), then also (mathbb{R} = {x-1} _{i=1}^{varphi (b)}).

**Proof****Statement 1:**The existence of a solution follows immediately from Be ́zout’s Lemma, namely(a′ = _{b} x) solves for (x) in (ax+by = 1). This solution must be in (mathbb{R}), because (a), in turn, is the solution of (a′x + by = 1) and thus Be ́zout’s Lemma implies that (gcd (a',b) = 1). Suppose we have two solutions (ax = _{b} 1) and (ay = _{b} 1), then uniqueness follows from applying the cancelation Theorem 2.7 to the difference of these equations.**Statement 2:**By hypothesis, (gcd (a,b) > 1). We have that (ax = _{b} 1) is equivalent to (ax+by = 1), which contradicts Be ́zout’s Lemma.**Statement 3:**This is similar to Lemma 5.3. By (1), we know that all inverses are in (R). So if the statement is false, there must be two elements of (R). So if the statement is false, there must be two elements of (R) with the same inverse: (ax = _{b} cx). This is impossible by cancelation.

Lemma 5.17

Let (p) be prime. Then (a^2 = _{p} 1) if and only if (a = _{p} pm 1)

**Proof**We have

[a^2 = _{p} 1 Leftrightarrow a^{2}-1 = _{p} (a+1)(a-1) = _{p} 0 Leftrightarrow p | (a+1)(a-1) onumber]

Because (p) is prime, Corollary 2.12 says that either (p | a+1) (and so (a = _{p} -1)) or (p | a-1) (and so (a = _{p} +1)).

Perhaps, surprisingly, this last lemma is false if (p) is not prime. For example, (4^2 = _{15} 1), but (4 e _{15} pm 1).

Theorem 5.18 (Wilson's theorem)

If (p) prime in (mathbb{Z}), then ((p-1)! = _{p} -1). If (b) is composite, then ((b-1)! e _{b} -1)

**Proof**This is true for (p = 2). If (p > 2), then Proposition 5.16(3) and Lemma 5.17 imply that every factor (a_{i}) in the product ((p-1)!) other than (-1) or (1) has a unique inverse (a_{i}') different from itself. The factors (a_{i}') run through all factors (2) through (p-2) exactly once. Thus in the product, we can pair each (a_{i}) different from (pm 1) with its inverse. This gives

[(p-1)! = _{p} (+1)(-1) prod a_{i}a_{i}' = _{p} -1 onumber]

The second part is easier. If (b) is composite, there are least residues (a) and (b) greater than (1) so that (ad = _{b} 0). Now either we can choose (a) and (b) distinct and then ((b-1)!) contains the product (ad), and thus it equals zero mod (b). Or else this is impssible and (b = a^2). But then still (gcd ((b-1)!, b) ge a). By Be ́zout, we must have ((b-1)!) mod (b) must be a multiple of (a).

Wilson's theorem could be used to test primality of a number (n). However, this takes (n) multiplications, which in practice is more expensive than trying to divide (n) by all numbers less than (sqrt{n}). Note, however, that if you want to compute a list of all primes between (1) and (N), Wilson’s theorem can be used much more efficiently. After computing ((k-1)! = _{k}) to determine whether (k) is prime, it takes only (1) multiplication and (1) division to determine whether (k+1) is prime.

Here is the take-away that will be important for Chapter **??**. More in particular, we have the following result.

Corollary 5.19

Let (p) be prime.

For every (a in mathbb{Z}_{p}) there is a unique (a′ = _{p} -a) such that (a+a′ = _{p} 0).

For every (a in mathbb{Z}_{p}) and (a e 0), there is a unique (a′ = a^{-1}) so that (aa′ = _{p} 1).

Addition and multiplication are well-defined in (mathbb{Z}_{b}) (see exercises 5.1 and 5.2). Thus when (p) is prime, we can add, multiply, subtract, and divide in (mathbb{Z}_{p}). In the words of Chapter **??**, when (p) is a prime, then (mathbb{Z}_{p}) is a field. It is an interesting fact that the same is not true for a composite number (b). According to Proposition 5.16, we need the reduced set of residues for the multiplication to be invertible. At the same time, for the set to be closed under multiplication, we need all of (mathbb{Z}_{b}) (think of (1+1+dots)). Thus the operations addition and multiplication in (mathbb{Z}_{b}) collaborate with each other only if (b) is a prime.

## Number of elements in Quotient ring over $mathbb Z_5[x]$ and $mathbb Z_[x]$

a) How many elements does the quotient ring$displaystyle frac

I can see that the polynomial, $displaystyle p(x)= x^2+1=(x-2)(x-3)$ is reducible over the field of integers modulo $5$ but can't proceed further.

Where the polynomial was irreducible over the field of integers modulo $11$.

I had a look at some solutions which say that the elements in this quotient ring will be of type $ax+b$ and then we have $11$ choices for each of the two and consequently $121$ elements.

I could not follow why the elements will be of $ax+b$ form. Please explain.

Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee, 247667, India

*** Corresponding author: Amit Sharma**

**Received** October 2017 **Revised** March 2018 **Published** September 2018

In this paper, we study a class of skew-cyclic codes using a skew polynomial ring over $R = mathbb

##### References:

M. Araya, M. Harada, H. Ito and K. Saito, On the classification of Z4-codes, *Adv. Math. Commun.*, **11** (2017), 747-756. doi: 10.3934/amc.2017054. Google Scholar

N. Aydin and T. Asamov, A database of Z4 codes, *J. Comb. Inf. Syst. Sci.*, **34** (2009), 1-12. Google Scholar

M. Bhaintwal, Skew quasi-cyclic codes over Galois rings, *Des. Codes Cryptogr.*, **62** (2012), 85-101. doi: 10.1007/s10623-011-9494-0. Google Scholar

I. F. Blake, Codes over certain rings, *Information and Control.*, **20** (1972), 396-404. doi: 10.1016/S0019-9958(72)90223-9. Google Scholar

I. F. Blake, Codes over integer residue rings, *Information and Control.*, **29** (1975), 295-300. doi: 10.1016/S0019-9958(75)80001-5. Google Scholar

W. Bosma, J. J. Cannon, C. Fieker and A. Steel, Handbook of magma functions, *Edition*, **2** (2010), 5017 pages. Google Scholar

D. Boucher and F. Ulmer, Coding with skew polynomial rings, *J. of Symbolic Comput.*, **44** (2009), 1644-1656. doi: 10.1016/j.jsc.2007.11.008. Google Scholar

D. Boucher, W. Geiselmann and F. Ulmer, Skew cyclic codes, *Appl. Algebra Engrg. Comm. Comput.*, **18** (2007), 379-389. doi: 10.1007/s00200-007-0043-z. Google Scholar

D. Boucher and F. Ulmer, Codes as modules over skew polynomial rings, *In Proc. of* 12 *th* *IMA International Conference, Cryptography and Coding, Cirencester, UK, LNCS*, **5921** (2009), 38-55. doi: 10.1007/978-3-642-10868-6_3. Google Scholar

D. Boucher, P. Sol$acute*Adv. Math. Commun.*, **2** (2008), 273-292. doi: 10.3934/amc.2008.2.273. Google Scholar

D. Boucher and F. Ulmer, Linear codes using skew polynomials with automorphisms and derivations, *Des. Codes Cryptogr.*, **70** (2014), 405-431. doi: 10.1007/s10623-012-9704-4. Google Scholar

S. T. Dougherty and K. Shiromoto, Maximum distance codes over rings of order 4, *IEEE Trans. Info Theory*, **47** (2001), 400-404. doi: 10.1109/18.904544. Google Scholar

F. Gursoy, I. Siap and B. Yildiz, Construction of skew cyclic codes over $mathbb*Adv. Math. Commum.*, **8** (2014), 313-322. doi: 10.3934/amc.2014.8.313. Google Scholar

Jr. A. R. Hammons, P. V. Kumar, A. R. Calderbank, N. J. Sloane and P. Sol$acute*IEEE Trans. Inform. Theory*, **40** (1994), 301-319. doi: 10.1109/18.312154. Google Scholar

S. Jitman, S. Ling and P. Udomkavanich, Skew constacyclic codes over finite chain rings, *Adv. Math. Commun.*, **6** (2012), 39-63. doi: 10.3934/amc.2012.6.39. Google Scholar

B. R. McDonald, *Finite Rings with Identity*, Marcel Dekker Inc, New York, 1974. Google Scholar

M. Ozen, F. Z. Uzekmek, N. Aydin and N. T. Ozzaim, Cyclic and some constacyclic codes over the ring $frac*Finite Fields Appl.*, **38** (2016), 27-39. doi: 10.1016/j.ffa.2015.12.003. Google Scholar

E. Prange, Cyclic error-correcting codes in two symbols, *Air Force Cambridge Research Center, Cambridge, MA, Tech. Rep. AFCRC-TN*, (1957), 57-103. Google Scholar

M. Shi, L. Qian, L. Sok, N. Aydin and P. Sole, On constacyclic codes over $frac*Finite Fields Appl.*, **45** (2017), 86-95. doi: 10.1016/j.ffa.2016.11.016. Google Scholar

I. Siap, T. Abualrub, N. Aydin and P. Seneviratne, Skew cyclic codes of arbitrary length, *Int. J. Inf. Coding Theory*, **2** (2011), 10-20. doi: 10.1504/IJICOT.2011.044674. Google Scholar

E. Spiegel, Codes over $mathbb*Information and Control.*, **35** (1977), 48-51. doi: 10.1016/S0019-9958(77)90526-5. Google Scholar

E. Spiegel, Codes over $mathbb*Information and Control.*, **37** (1978), 100-104. doi: 10.1016/S0019-9958(78)90461-8. Google Scholar

B. Yildiz and N. Aydin, On codes over $mathbb*Int. J. Inf. Coding Theory*, **2** (2014), 226-237. doi: 10.1504/IJICOT.2014.066107. Google Scholar

B. Yildiz and S. Karadeniz, Linear codes over $mathbb*Finite Fields Appl.*, **27** (2014), 24-40. doi: 10.1016/j.ffa.2013.12.007. Google Scholar

##### References:

M. Araya, M. Harada, H. Ito and K. Saito, On the classification of Z4-codes, *Adv. Math. Commun.*, **11** (2017), 747-756. doi: 10.3934/amc.2017054. Google Scholar

N. Aydin and T. Asamov, A database of Z4 codes, *J. Comb. Inf. Syst. Sci.*, **34** (2009), 1-12. Google Scholar

M. Bhaintwal, Skew quasi-cyclic codes over Galois rings, *Des. Codes Cryptogr.*, **62** (2012), 85-101. doi: 10.1007/s10623-011-9494-0. Google Scholar

I. F. Blake, Codes over certain rings, *Information and Control.*, **20** (1972), 396-404. doi: 10.1016/S0019-9958(72)90223-9. Google Scholar

I. F. Blake, Codes over integer residue rings, *Information and Control.*, **29** (1975), 295-300. doi: 10.1016/S0019-9958(75)80001-5. Google Scholar

W. Bosma, J. J. Cannon, C. Fieker and A. Steel, Handbook of magma functions, *Edition*, **2** (2010), 5017 pages. Google Scholar

D. Boucher and F. Ulmer, Coding with skew polynomial rings, *J. of Symbolic Comput.*, **44** (2009), 1644-1656. doi: 10.1016/j.jsc.2007.11.008. Google Scholar

D. Boucher, W. Geiselmann and F. Ulmer, Skew cyclic codes, *Appl. Algebra Engrg. Comm. Comput.*, **18** (2007), 379-389. doi: 10.1007/s00200-007-0043-z. Google Scholar

D. Boucher and F. Ulmer, Codes as modules over skew polynomial rings, *In Proc. of* 12 *th* *IMA International Conference, Cryptography and Coding, Cirencester, UK, LNCS*, **5921** (2009), 38-55. doi: 10.1007/978-3-642-10868-6_3. Google Scholar

D. Boucher, P. Sol$acute*Adv. Math. Commun.*, **2** (2008), 273-292. doi: 10.3934/amc.2008.2.273. Google Scholar

D. Boucher and F. Ulmer, Linear codes using skew polynomials with automorphisms and derivations, *Des. Codes Cryptogr.*, **70** (2014), 405-431. doi: 10.1007/s10623-012-9704-4. Google Scholar

S. T. Dougherty and K. Shiromoto, Maximum distance codes over rings of order 4, *IEEE Trans. Info Theory*, **47** (2001), 400-404. doi: 10.1109/18.904544. Google Scholar

F. Gursoy, I. Siap and B. Yildiz, Construction of skew cyclic codes over $mathbb*Adv. Math. Commum.*, **8** (2014), 313-322. doi: 10.3934/amc.2014.8.313. Google Scholar

Jr. A. R. Hammons, P. V. Kumar, A. R. Calderbank, N. J. Sloane and P. Sol$acute*IEEE Trans. Inform. Theory*, **40** (1994), 301-319. doi: 10.1109/18.312154. Google Scholar

S. Jitman, S. Ling and P. Udomkavanich, Skew constacyclic codes over finite chain rings, *Adv. Math. Commun.*, **6** (2012), 39-63. doi: 10.3934/amc.2012.6.39. Google Scholar

B. R. McDonald, *Finite Rings with Identity*, Marcel Dekker Inc, New York, 1974. Google Scholar

M. Ozen, F. Z. Uzekmek, N. Aydin and N. T. Ozzaim, Cyclic and some constacyclic codes over the ring $frac*Finite Fields Appl.*, **38** (2016), 27-39. doi: 10.1016/j.ffa.2015.12.003. Google Scholar

E. Prange, Cyclic error-correcting codes in two symbols, *Air Force Cambridge Research Center, Cambridge, MA, Tech. Rep. AFCRC-TN*, (1957), 57-103. Google Scholar

M. Shi, L. Qian, L. Sok, N. Aydin and P. Sole, On constacyclic codes over $frac*Finite Fields Appl.*, **45** (2017), 86-95. doi: 10.1016/j.ffa.2016.11.016. Google Scholar

I. Siap, T. Abualrub, N. Aydin and P. Seneviratne, Skew cyclic codes of arbitrary length, *Int. J. Inf. Coding Theory*, **2** (2011), 10-20. doi: 10.1504/IJICOT.2011.044674. Google Scholar

E. Spiegel, Codes over $mathbb*Information and Control.*, **35** (1977), 48-51. doi: 10.1016/S0019-9958(77)90526-5. Google Scholar

E. Spiegel, Codes over $mathbb*Information and Control.*, **37** (1978), 100-104. doi: 10.1016/S0019-9958(78)90461-8. Google Scholar

B. Yildiz and N. Aydin, On codes over $mathbb*Int. J. Inf. Coding Theory*, **2** (2014), 226-237. doi: 10.1504/IJICOT.2014.066107. Google Scholar

B. Yildiz and S. Karadeniz, Linear codes over $mathbb*Finite Fields Appl.*, **27** (2014), 24-40. doi: 10.1016/j.ffa.2013.12.007. Google Scholar

Download as PowerPoint slide

$C$ | $Phi(C)$ | $Res(C)$ | $C^*$ | |

Set of generators | Code | $(n, 4^ | $(n, 4^ | $(n, 4^ |

$ | $C_1$ | $<(10, 4^6, 2)^<>>$ | $<(5, 4^42^1, 2)^<*>>$ | $mathbf<(10, 4^82^2, 2)>^<**>$ |

$ | $C_2$ | $<(20, 4^6, 8)>$ | $(10, 4^6, 4)^*$ | $(20, 4^<12>, 4)^*$ |

$ | $C_3$ | $<(20, 4^6, 6)>$ | $(10, 4^5, 6)^*$ | $(20, 4^<10>, 6)$ |

$left< < | $C_4$ | $<(24, 4^8, 6)>$ | $(12, 4^8, 4)^*$ | $(24, 4^<16>, 4)^*$ |

$left< < | $C_5$ | $<(28, 4^8, 6)>$ | $(14, 4^8, 5)^*$ | $(28, 4^<16>, 5)^*$ |

$left< < | $C_6$ | $<(30, 4^8, 6)>$ | $(15, 4^8, 6)^*$ | $(30, 4^<16>, 6)$ |

$left< < | $C_7$ | $<(36, 4^8, 8)>$ | $(18, 4^8, 8)^*$ | $(36, 4^<16>, 8)^*$ |

$C$ | $Phi(C)$ | $Res(C)$ | $C^*$ | |

Set of generators | Code | $(n, 4^ | $(n, 4^ | $(n, 4^ |

$ | $C_1$ | $<(10, 4^6, 2)^<>>$ | $<(5, 4^42^1, 2)^<*>>$ | $mathbf<(10, 4^82^2, 2)>^<**>$ |

$ | $C_2$ | $<(20, 4^6, 8)>$ | $(10, 4^6, 4)^*$ | $(20, 4^<12>, 4)^*$ |

$ | $C_3$ | $<(20, 4^6, 6)>$ | $(10, 4^5, 6)^*$ | $(20, 4^<10>, 6)$ |

$left< < | $C_4$ | $<(24, 4^8, 6)>$ | $(12, 4^8, 4)^*$ | $(24, 4^<16>, 4)^*$ |

$left< < | $C_5$ | $<(28, 4^8, 6)>$ | $(14, 4^8, 5)^*$ | $(28, 4^<16>, 5)^*$ |

$left< < | $C_6$ | $<(30, 4^8, 6)>$ | $(15, 4^8, 6)^*$ | $(30, 4^<16>, 6)$ |

$left< < | $C_7$ | $<(36, 4^8, 8)>$ | $(18, 4^8, 8)^*$ | $(36, 4^<16>, 8)^*$ |

$C$ | $Phi(C)$ | $Res(C)$ | $C^*$ | |

Set of generators | Name | $(n, M, d_L)$ | $(n, 4^ | $(n, 4^ |

$ | $A_1$ | $<(10,128, 2)^<>>$ | $<(5, 4^32^1, 2)^<*>>$ | $(10, 4^62^2, 2)$ |

$ | $A_2$ | $<(12, 4096, 2)^<>>$ | $<(6, 4^52^1, 2)^<*>>$ | $mathbf<(12, 4^<10>2^2, 2)>^<**>$ |

$ | $A_3$ | $<(14, 65536, 2)^<>>$ | $<(7, 4^62^1, 2)^<*>>$ | $<(14, 4^<12>2^2, 2)>$ |

$ | $A_3$ | $<(16, 65536, 4)^<>>$ | $mathbf<(8, 4^7, 2)^<**>>$ | $mathbf<(16, 4^<14>, 2)>^<**>$ |

$C$ | $Phi(C)$ | $Res(C)$ | $C^*$ | |

Set of generators | Name | $(n, M, d_L)$ | $(n, 4^ | $(n, 4^ |

$ | $A_1$ | $<(10,128, 2)^<>>$ | $<(5, 4^32^1, 2)^<*>>$ | $(10, 4^62^2, 2)$ |

$ | $A_2$ | $<(12, 4096, 2)^<>>$ | $<(6, 4^52^1, 2)^<*>>$ | $mathbf<(12, 4^<10>2^2, 2)>^<**>$ |

$ | $A_3$ | $<(14, 65536, 2)^<>>$ | $<(7, 4^62^1, 2)^<*>>$ | $<(14, 4^<12>2^2, 2)>$ |

$ | $A_3$ | $<(16, 65536, 4)^<>>$ | $mathbf<(8, 4^7, 2)^<**>>$ | $mathbf<(16, 4^<14>, 2)>^<**>$ |

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To make a number change to the next lower whole number (integer), get the number’s *floor* value. The floor value for 8.76 is 8 since that is the next lower whole number. For the negative number of 6.17 , its floor is -7 since that’s the next lower whole number.

The fractional part of a number is removed by *truncating* it. If a number has the value 54.234 its truncated value is 54 . Truncation works the same way for a negative number. The truncated value of -34.913 is -34 .

The simplest examples of abelian groups are **cyclic groups**, which are groups generated by a single element and thus isomorphic to Z n mathbb

Though all cyclic groups are abelian, not all abelian groups are cyclic. For instance, the Klein four group Z 2 × Z 2 mathbb

In contrast, the group of invertible matrices with a group law of matrix multiplication do not form an abelian group (it is **nonabelian**), because it is not generally true that M N = N M MN = NM M N = N M for matrices M , N M,N M , N . The symmetric group S n S_n S n is also nonabelian for n ≥ 3 n geq 3 n ≥ 3 .

Rings are also examples of abelian groups, with respect to their additive operations. Further, the units of a ring form an abelian group with respect to its multiplicative operation. For example, the real numbers form an additive abelian group, and the nonzero real numbers (denoted R ∗ mathbb

## MAT 112 Ancient and Contemporary Mathematics

In Definition 1.3.10 we had defined multiplication as repeated addition. For positive integers we conducted division as repeated subtraction. We first consider this case and then generalize the algorithm to all integers by giving a division algorithm for negative integers.

Watch the video in Figure 3.2.1 on the Division algorithm and then read the detailed description in the remainder of this section.

### Subsection 3.2.1 Division Algorithm for positive integers

In our first version of the division algorithm we start with a non-negative integer (a) and keep subtracting a natural number (b) until we end up with a number that is less than (b) and greater than or equal to (0 ext<.>) We call the number of times that we can subtract (b) from (a) the of the division of (a) by (b ext<.>) The remaining number is called the of the division of (a) by (b ext<.>)

We typically use the variable (q) for the quotient and the variable (r) for the remainder. We have

The division algorithm computes the quotient as well as the remainder. In Algorithm 3.2.2 and Algorithm 3.2.10 we indicate this by giving two values separated by a comma after the *return*.

If (alt b) then we cannot subtract (b) from (a) and end up with a number greater than or equal to (b ext<.>) Thus, in this case, the quotient is 0 and the remainder is (a) itself. We catch this case in step 1 of the algorithm.

###### Algorithm 3.2.2 . Division for positive numbers.

a natural number (a) and a natural number (b)

Two integers (q) and (r) such that (a=(qcdot b)+r) and (0 leq rlt b)

We first consider an example in which the algorithm terminates before we enter the *repeat_until* loop.

###### Example 3.2.3 . Dividing (4) by (7) with Algorithm 3.2.2.

We find the output values of Algorithm 3.2.2 for the input values (a=4) and (b=7 ext<.>)

As (a=4) and (b=7) statement (a lt b) is true. So we follow the instruction after *then* and return the values of (q) and (r ext<,>) namely 0 and 4.

Thus the quotient of the division of (4) by (7) is (0) and the remainder is (4 ext<.>)

With the division algorithm we find a quotient and remainder. In this example we go through the *repeat_until* loop several times.

###### Example 3.2.4 . Dividing (30) by (8) with Algorithm 3.2.2.

We find the output values of the Algorithm 3.2.2 for the input values (a=30) and (b=8 ext<.>)

1. As (a=30) and (b=8) the statement (a lt b) is false. So we continue with step 2.

5. As (r=22) and (q=1) the statement (r lt q) is false. So we continue with step 4

5. As (r=14) and (q=8) the statement (r lt q) is false. So we continue with step 4

As (r=6) and (q=8) the statement (rlt q) is true. So we continue with step 6.

We return the quotient (q=3) and the remainder (r=6)

Thus the quotient of the division of (30) by (8) is (3) and the remainder is (6 ext<.>)

When working through the instructions of Algorithm 3.2.2 it can be more convenient to give the values of all relevant variables in each iteration of the loop in a table. We revisit Example 3.2.4 present the work in a more compact form.

###### Example 3.2.5 . Dividing (30) by (8) with Algorithm 3.2.2 shorter.

We find the output values of Algorithm 3.2.2 for the input values (a=30) and (b=8 ext<.>)

In each row of the table we write the values of all variables for an iteration of the loop. If a variable has no value we leave the entry blank. Similarly in for the output we leave the entries of all variables that are not part of the output blank.

steps | (a) | (b) | (q) | (r) |

Input | (30) | (8) | () | () |

1.,2.,3. | (30) | (8) | (0) | (30) |

4.,5. | (30) | (8) | (0+1=1) | (30-8=22) |

4.,5. | (30) | (8) | (1+1=2) | (22-8=14) |

4.,5. | (30) | (8) | (1+1=3) | (14-8=6) |

Output | () | () | (3) | (6) |

So the output is (q=3) and (r=6)

In Example 3.2.6 you can observe how the values of the variables change as you click your way through the steps of the division algorithm.

###### Example 3.2.6 . Division algorithm interactive.

In Checkpoint 3.2.7 we unroll the loop in Algorithm 3.2.2 in a similar way. Follow the instructions to find quotient and remainder.

###### Checkpoint 3.2.7 . Find quotient and remainder with division algorithm.

Sometimes one is not interested in both the quotient and the remainder. In such a case one can use a simplified algorithm. Determine the output of the algorithm in Checkpoint 3.2.8.

###### Checkpoint 3.2.8 . Another variant of the division algorithm.

If (a>0 ext<,>) then Algorithm 3.2.2 returns the quotient and remainder of the division of (a) by (b ext<.>) If we try to use Algorithm 3.2.2 when (a) is negative, the algorithm always returns (0,a) which does not satisfy the condition (0le r) for the output since (r=alt 0 ext<.>) So we need a different algorithm for the case (alt 0 ext<.>)

### Subsection 3.2.2 Division Algorithm for Negative Integers

When (alt 0 ext<,>) we still want find (q) and (r) such that (a=(qcdot b)+r) with (0le rlt b ext<.>) We get a positive remainder when (a) is negative by repeated addition of (b ext<.>) This is the same as repeatedly adding (-b ext<.>) Let (s) be the number of times we have to add (b) to (a) in order to get (0 le r lt b ext<.>) After (s) additions of (b) to (a) we have

If we let (q:=-s ext<,>) we get (r=a-(qcdot b)) (compare this to what we wanted). We stop when (0le rlt b ext<.>) We repeatedly add (b) to negative numbers until (0le rlt b) is true. Since a negative number plus (b) is always less than (b) and we check the value of (r) after every addition, it is sufficient to check whether (0le r ext<.>)

###### Example 3.2.9 . Dividing (-33) by (9).

We illustrate the process of dividing a negative number by dividing (-33) by (9 ext<.>) We repeatedly add (9) until we get a number from (0) to (9-1=8 ext<.>) That number is the remainder. The negative of the number of times we add (9) is the quotient.

As (0le 3lt 9) we are done. The remainder is (3 ext<.>) We have added (9) four times, so the quotient is (-4 ext<.>) We have

We now formalize this procedure in an algorithm.

###### Algorithm 3.2.10 . Division for negative integers.

A negative integer (a) and a natural number (b)

Two integers (q) and (r) such that (a=(qcdot b)+r) and (0 leq rlt b)

## How to Write a Division Sentence

**To write a division sentence use the following steps:**

** **

- Write down the total number being shared or divided first.
- Next write the division sign, ÷.
- After the division sign, write down the number of groups the amount is being shared into.
- Next write the equals sign, =.
- Finally, write the number in each group after the objects have been shared.

As long as we use whole numbers, the largest number in a division sentence will come first.

Here is an example of writing a division sentence for a word problem.

Ten marbles are put into 5 bags.

The first step is to write the total number being shared, which is 10.

The second step is to write a division sign, ÷.

The third step is to write the amount of groups. This is the number of bags the marbles will be put into. We have 5.

The fourth step is two write an equals sign, =.

The fifth step is to write the number in each group. By placing the marbles into 5 equal groups we can see that there are 5 in each group.

The division sentence is 10 ÷ 5 = 2.

This sentence means that 10 shared into 5 equal groups gives us 2 in each group.

Here is another example of writing a division sentence for a word problem.

12 apples are shared between 4 children.

Remember that the largest number in the division sentence will come first.

We are sharing 12 objects between 4 people.

Each child gets 3 and so, our answer to the division is 3.

12 ÷ 4 = 3 means that 12 apples shared between 4 children gives each child 3 apples.

Here is an example of writing a division.

16 birds are put into groups of 4.

The largest number is the total. We have 16 birds so we write this first.

The number after the division sign is the number of groups.

We have 16 ÷ 4, which means 16 birds shared into 4 equal groups.

We count how many birds in each group to get our answer, after the equals sign.

There are 4 birds in each group and so, 16 ÷ 4 = 4.

Now try our lesson on *Short Division without Remainders* where we learn how to use the short division method for dividing numbers.

## Simple Division Word Problems

To solve a division word problem, we can use the following steps:

- Identify the numbers given in the question.
- Identify which number is the total quantity.
- Identify how many groups we are sharing between or how many need to go in each group.
- Divide the total by the number of groups to find the amount in each group.
- Or divide the total by how many needed in each group to find out how many groups can be made.

In this example, ‘I have 10 sweets to **share evenly** between 5 children. How many sweets do they **each** get?’

We are trying to find how many sweets each child gets, so we want to know how many sweets will be in each group.

We first identify the total, which is 10 sweets.

Now we identify how many groups we have, which is 5. We are sharing equally between 5 children.

We divide the total amount by the number of groups.

We are sharing 10 sweets between 5 people.

10 ÷ 5 = 2 and so, each child gets 2 sweets each.

When teaching division word problems, we can draw 10 sweets and group them equally by drawing circles around them to help visualise this. We could also get 10 counters and share them out equally, one at a time.

We can see in this example that we had the keywords of **share evenly** and **each**, which can give us a clue that we have a division.

The division also tells us how many times 5 goes into 10.

In this next example, ‘I have 80 matches. I will put 8 into each packet. How any packets will I fill?’

We want to see how many packets we will fill. We want to see how many groups we will create.

We first identify the total number of matches, which is 80.

We then identify the number in each packet, which is 8.

To find the total number of packets, we will divide.

80 ÷ 8 = 10 and so, we can make 10 packets.

We can think of this as working out how many times 8 goes into 80 or how many packets can be made from 80 matches.

In this next example, ‘I need 30 crayons. Each pack contains 5 crayons. How many packs should I buy?’

The total number is the larger number, which is is 30.

We are buying the crayons in equal groups of 5.

We need to work out how many groups we need. How many fives make 30?

We need to work out how many fives go into 30.

30 ÷ 5 = 6 and so, we need 6 packs.

We can check out answer. 6 lots of 5 make the 30 crayons needed because 6 × 5 = 30.

In this example we needed to find the number of groups required. So we divided the total by the number in each group.

In this example, ‘I have 21 chairs. I will arrange the chairs in rows of 7.How many rows should I make?’

Here we have the total number of chairs, which is 21.

We are arranging them into rows of 7, so each group contains 7 chairs.

We want to find the number of rows that we can make. We want to work out how many rows of 7 can be made from 21 chairs. This is how many times 7 goes into 21.

21 ÷ 7 = 3 and so, we can make 3 rows.

We can see that each row is the same size. We can teach this by taking 21 counters and sharing them into 3 equal rows.

In this example involving money, ‘Shirts cost $11 and I have $66. How many shirts can I buy?’

We want to know how many elevens go into 66.

The total is $66 and we are dividing by 11.

We want to know how many times we can spend $11.

66 ÷ 11 = 6 and so, we can spend $11 six times.

Now try our lesson on *Short Division without Remainders* where we learn how to use the short division method to divide numbers.