Algebraic Reasoning (with Solved Exercises)

3412
Charles McCarthy

The algebraic reasoning It essentially consists of communicating a mathematical argument through a special language, which makes it more rigorous and general, making use of algebraic variables and operations defined among themselves. A characteristic of mathematics is the logical rigor and abstract tendency used in its arguments..

For this it is necessary to know the correct "grammar" to use in this writing. Furthermore, algebraic reasoning avoids ambiguities in the justification of a mathematical argument, which is essential to prove any result in mathematics..

Article index

  • 1 Algebraic variables
  • 2 Algebraic expressions
    • 2.1 Examples
  • 3 Solved exercises
    • 3.1 First exercise
    • 3.2 Second exercise
    • 3.3 Third exercise
  • 4 References

Algebraic variables

An algebraic variable is simply a variable (a letter or symbol) that represents a certain mathematical object..

For example, the letters x, y, z, are often used to represent the numbers that satisfy a given equation; the letters p, q r, to represent propositional formulas (or their respective capital letters to represent specific propositions); and the letters A, B, X, etc., to represent sets.

The term "variable" emphasizes that the object in question is not fixed, but varies. Such is the case of an equation, in which variables are used to determine solutions that are unknown in principle..

In general terms, an algebraic variable can be considered as a letter that represents some object, whether it is fixed or not..

Just as algebraic variables are used to represent mathematical objects, we can also consider symbols to represent mathematical operations.

For example, the symbol "+" represents the operation "addition." Other examples are the different symbolic notations of logical connectives in the case of propositions and sets..

Algebraic expressions

An algebraic expression is a combination of algebraic variables through previously defined operations. Examples of this are the basic operations of addition, subtraction, multiplication and division between numbers, or the logical connectives in propositions and sets.

Algebraic reasoning is responsible for expressing a mathematical reasoning or argument through algebraic expressions.

This form of expression helps to simplify and abbreviate the writing, since it makes use of symbolic notations and allows a better understanding of the reasoning, presenting it in a clearer and more precise way..

Examples

Let's look at some examples that show how algebraic reasoning is used. It is used very regularly to solve logic and reasoning problems, as we will see shortly..

Consider the well-known mathematical proposition "the sum of two numbers is commutative." Let's see how we can express this proposition algebraically: given two numbers "a" and "b", what this proposition means is that a + b = b + a.

The reasoning used to interpret the initial proposition and express it in algebraic terms is algebraic reasoning..

We could also mention the famous expression "the order of the factors does not alter the product", which refers to the fact that the product of two numbers is also commutative, and is algebraically expressed as axb = bxa.

Similarly, the associative and distributive properties for addition and product, in which subtraction and division are included, can be (and in fact are) expressed algebraically..

This type of reasoning encompasses a very broad language and is used in many different contexts. Depending on each case, in these contexts it is necessary to recognize patterns, interpret sentences and generalize and formalize their expression in algebraic terms, providing valid and sequential reasoning..

Solved exercises

The following are some logic problems, which we will solve using algebraic reasoning:

First exercise

What is the number that, taking half of it, is equal to one?

Solution

To solve this type of exercise, it is very useful to represent the value that we want to determine by means of a variable. In this case, we want to find a number that, when taking half of it, gives the number one as a result. Let us denote by x the number sought.

"Taking half" from a number implies dividing it by 2. So the above can be expressed algebraically as x / 2 = 1, and the problem boils down to solving an equation, which in this case is linear and very easy to solve. Solving for x we ​​obtain that the solution is x = 2.

In conclusion, 2 is the number that when taking half is equal to 1.

Second exercise

How many minutes until midnight if 10 minutes ago 5/3 of what is left now?

Solution

Let us denote by "z" the number of minutes until midnight (any other letter can be used). That is to say that right now there are “z” minutes until midnight. This implies that 10 minutes ago “z + 10” minutes were missing for midnight, and this corresponds to 5/3 of what is missing now; that is, (5/3) z.

Then the problem boils down to solving the equation z + 10 = (5/3) z. Multiplying both sides of the equality by 3, we obtain the equation 3z + 30 = 5z.

Now, when grouping the variable "z" on one side of equality, we obtain that 2z = 15, which implies that z = 15.

So it's 15 minutes to midnight.

Third exercise

In a tribe that practice bartering, there are these equivalences:

- A spear and a necklace are exchanged for a shield.

- A spear is equivalent to a knife and a necklace.

- Two shields are exchanged for three units of knives.

How many necklaces is a spear equivalent to?

Solution

Sean:

Co = a necklace

L = a spear

E = a shield

Cu = a knife

So we have the following relationships:

Co + L = E

L = Co + Cu

2E = 3Cu

So the problem boils down to solving a system of equations. Despite having more unknowns than equations, this system can be solved, since they do not ask us for a specific solution but rather one of the variables as a function of another. What we must do is express "Co" in terms of "L" exclusively.

From the second equation we have that Cu = L - Co. Substituting in the third one we obtain that E = (3L - 3Co) / 2. Finally, substituting in the first equation and simplifying it is obtained that 5Co = L; that is, a spear is equal to five necklaces.

References

  1. Billstein, R., Libeskind, S., & Lott, J. W. (2013). Mathematics: A Problem Solving Approach for Elementary Education Teachers. López Mateos Editors.
  2. Fuentes, A. (2016). BASIC MATH. An Introduction to Calculus. Lulu.com.
  3. García Rua, J., & Martínez Sánchez, J. M. (1997). Elementary basic mathematics. Ministry of Education.
  4. Rees, P. K. (1986). Algebra. Reverte.
  5. Rock, N. M. (2006). Algebra I Is Easy! So easy. Team Rock Press.
  6. Smith, S. A. (2000). Algebra. Pearson Education.
  7. Szecsei, D. (2006). Basic Math and Pre-Algebra (illustrated ed.). Career Press.

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