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Rings in High School Mathematics

STUDYING LINEAR EQUATIONS OF TYPE ax + b = c VIA RINGS

Prerequisite: Knowledge of rings.

INTRODUCTION/SUMMARY

Mathematical taxonomy allows us to get as precise as we want to. So, when we speak of rings, we also want to move up the taxonomicalladder as we desire. We may move up to integraldomain, and, finally, get to field. Clearly integral domains and fields are rings. We willstudy the linear equation ax + b = c, under each of these sets, successively.

Solving ax + b = c in the Integers

The set of integers with addition (+) and multiplication (*) is an integral domain. A discussion and some definitions are now in order.

Discussion. O. (0 Divisors). Consider 2 * 3 in (Z6, +, *). Cleary, here, 2 * 3 = 0. It then follows that 2 and 3 are both 0 divisors inZ6 . On, the other hand, a * b = 0 for a, b in Z, implies that a orb (or both are 0). For example the equation 2 * b = 0 in Z clearly implies thatb cannotbe nonzero. That is, Z does not have 0 divisors.

Definition. R 1. Let a, and b belong to a ring a ring R. Then a b = a + -b.We call subtraction. Clearly, if a, b belongs to R, then a b belongs to Ras well.Definition. ID 1(Integral Domain) An integral domain is a commutative ring with unity (multiplicative identity element) 1 0 and no 0divisors.(Z, +, *) is an integral domain that concerns us in this paper. Clearly, the rationals and the reals with + and * are also integral domains.

We now discuss solving ax + b = c in the integers, Z.

Proposition. ID 2. Let a, b, and c be integers such that a 0. Then, the solution to ax + b = c is an integer if and only if a(c b).

Proof

Let a, b, and c be integers such that a 0. And let the solution to ax + b = c be an integer. Then x = (c b)/a is an integer whence a(c b).For the other direction, let a(c b). Then by (c b)/a must be an integer.

Example 1

5x + 8 = 23 has an integer solution since 5(23 8). Clearly, x = (23 8)/5 = 3.

Example 25x + 8 = 22 does not have an integer solution since 5 does not divide (22 8). That is, (22 8)/5 = 14/5 is not an integer.We encounter problems like example 2 in high school algebra. But, sadly some students might not understand the implications

ofproposition ID 2

. So, they may get a problem like example 2 and not know what the solution implies. Or worse, they may not know whatkind of solution to expect. We now consider what I believe to be the implication(s).

Discussion. F. (Fields). We know that a field is an integral domain that is endowed with some additional properties. In addition to all theproperties of integral domain holding in a field F, we must also have that (1) F {0} is closed under division, (2) if x 0, 0/x = 0 belongs to F,(3) every nonzero element in F has a unique multiplicative inverse in F.

From our discussion above, it is thus clear why a field is also called a commutative division ring. We now revisit the linear equation ax + b= c.Solving ax + b = c in the Rationals

It should be immediately obvious that if a, b, and c are integers or rationals, we can find an integer or a rational solution. That is, we do notneed to extend our discussion to the set of real numbers.

We now consider a proposition.

Proposition. F1. Let a, b, and c belong to the field of rational numbers, Q. Then the solution toax + b = c is rational.

ProofLet a, b, and c belong to Q such that a 0, and let ax + b = c. Then, x = (c b)/a. Butc b = c + -b is rational, by definition R1 (and closure property). And thus, from discussion F,(c b)/a is rational.

We reconsider example 2, and we see that, by proposition R1, 5x + 8 = 22 must have rational solution.

Example 3

Consider (1/2)x + 2/7 = 3/5. We see that , 2/7, and 3/5 are rational numbers, and 0.Clearly, x = [3/5 2/7]/(1/2)= 22/35 which is a rational number.

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7/28/2019 Rings in High School Mathematics

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If we consider ax + b = cx + d where a 0, b, c, and d are rationals with the caveat that a c, we can also apply our discussion to this type ofproblem.Just to jolt a mathematical nerve, we may want to ask our school students what would happen if 0 a = c while b d in the equation ax + b =cx + d. (We know the answer.)

In conclusion: This paper encourages us to encourage school students learn what to expect when we are solving linear equations of thetype ax + b = c. Often, by knowing what to look for (by having a mathematical hunch about what could be expected), a student may besaved a lot of unnecessary headaches when solving equations. We just discussed a simple type of equation (ax + b = c), but the spirit of thediscussion is pervasive.