Understanding Algebra for Calculus

Subsection 2.2.1 Inequivalent Expressions

Two expressions are not equivalent if there are some values of the variables in the expression that yield different numbers when you substitute. So, if you plug the same number into two expressions and get different results, then the two expressions are inequivalent (assuming you did the arithmetic correctly). With practice, you’ll get good at being able to detect inequivalent expressions without even needing to substitute anything. For now, we’ll look at some important examples that sometimes trip people up.

The two expressions \((x+y)^2\) and \(x^2 + y^2\) are not equivalent. One may detect this by noting the order in which operations are performed on \(x\) and \(y\text{.}\) In one expression you add the two variables, then square. In the other, you square each variable, then add. Based one our experience with order of operations (order matters), we think these may not be equivalent. It turns out that if you pick any two numbers for \(x\) and \(y\text{,}\) as long as neither is zero, these two expressions will evaluate to different numbers. For instance, if we substitute \(x=1\) and \(y=2\text{,}\) then the first expression gives

\begin{equation*} (1+2)^2 = 3^2 =9, \end{equation*}

whereas the second expression gives

\begin{equation*} 1^2+2^2 = 1+4 = 5. \end{equation*}

It is also possible to see that these two expressions are inequivalent by giving the geometric representations. We begin by thinking of \(x\) and \(y\) as lengths. Then \((x+y)^2\) is the area of a square whose side length is \(x+y\text{,}\) while \(x^2+y^2\) is the sum of areas of squares that fit inside the square of side length \(x+y\) as in the following diagram:

In this figure we see that the total shaded area is \(x^2+y^2\) while the area of the entire square is \((x+y)^2\text{.}\) If you note that the unshaded portions of the square each have area \(xy\text{,}\) we can see that \((x+y)^2\) is equivalent to \(x^2+2xy+y^2\) (That can be seen symbollically using the distributive law 2.2.19, which we will see soon.).

Subsection 2.2.2 Reordering and Regrouping

Sometimes the order in which operations are performed in an algebraic expression doesn't affect the result. Specifically, if the expression only involves addition, order of operations doesn't tell you which one to do first because it doesn't matter. Likewise for multiplication. Specifically, we have the following rules for addition and multiplication:

In Chapter 1 you were asked to explain the commutativity of addition geometrically using lengths on the number line (1.1.5). Explain the commutativity of multiplication, by thinking of \(a\) and \(b\) as lengths of the sides of a rectangle similar to the way we thought of \(x\) and \(y\) in 2.2.4.

Solution.

Using the fact that the area of a rectangle is given by

\begin{equation*} \mbox{Area} = \mbox{base}\times\mbox{height}, \end{equation*}

we can arrange our rectangle with \(a\) on the bottom and \(b\) on the side to get the area to be \(ab\text{.}\) If we simply rotate the rectangle \(90\) degrees, the are is then seen as \(ba\text{.}\) Since rotation doesn't change the area, we must have \(ab = ba\text{.}\)

These rules are most useful in simplifying complex expressions that involve many terms being added together or multiplied.

In the last step of the previous example we performed an important type of regrouping known as combining like terms. This is when all the terms with the same algebraic type (same exponents on a variable for example) are combined into a single term by adding together the constants by which they are multiplied. It is important to note that terms can only be combined when the only difference between them is a constant multiplier. For instance,

The remaining terms cannot be combined further because one has an exponent of \(2\) while the other has an exponent of \(\frac{1}{2}\text{.}\)

Subsection 2.2.3 Distributing, Expanding, and Factoring

One of the most important rules in constructing equivalent algebraic expressions is the Distributive Law.

This fact asserts that you can

• add two lengths, \(b\) and \(c\text{,}\) on the line together, then expand/compress by a factor of \(a\text{,}\) or

• expand/contract each of \(b\) and \(c\) by a factor of \(a\text{,}\) then add the results together.

The end result is the same in either case because the stretch/compression caused by multiplication by \(a\) is uniformly applied to all lengths. We can also see this in terms of areas like we did with other properties of the real numbers that involve multiplication (like 2.2.4 again). In the equation

We can think of \(b\) and \(c\) as horizontal lengths and \(a\) as a vertical length as in the diagram below:

When we glue these two recangles together, we have one rectangle whose area is \(a(b+c).\) Since gluing didn't change the total area, we have

Many seemingly complex algebraic manipulations boil down to using the distributive law several times and collecting like terms; it's just that \(a\text{,}\) \(b\text{,}\) and \(c\) may be some sort of algebraic expressions.

Now is probably a good time to just get a little practice expanding/distributing and collecting like terms. Just remember the distributive law and you'll see it's not that hard.

Now we will go backward from expanding/distributing. This process is known as factoring. When we expand an expression, we write it as an equivalent expression that is a sum of simple terms that each only involve multiplication and/or division. To factor an expression means to write it as a product of simple terms. For example, earlier in this chapter we had two expressions for the temperature in \(^{\circ}\)C in terms of the temperature in \(^{\circ}\)F. The expression \(C = \frac{5}{9}F - \frac{160}{9}\) is in expanded form, while \(C = \frac{5}{9}(F-32)\) is in factored form. Factored forms of expressions are often useful for solving equations as we'll see in the next chapter. Factoring isn't something that can be done in all cases by a single procedure; the following are just a few examples of common factoring methods.

The second solution is the so-called \(ac\) method including the reasoning behind it. Applying it blindly isn't terribly insightful in terms of explaining what factoring really does, but it gets the answer. For this reason it is recorded below:

Example 2.25. Any expression of the form \(a^2-b^2\) may be factored as \((a+b)(a-b)\) (\(a\) and \(b\) may represent more complicated expressions in this). One may check this using the distributive law. To see how this factorization works geometrically, consider the following figure with \(b\) being thought of as a number smaller than \(a\text{.}\)

In this figure the entire square has an area of \(a^2\) units and the gray shaded area has an area of \(a^2-b^2\) units. Each of the smaller shaded rectangles has area \(b(a-b)\) and the lower right shaded square has an area of \((a-b)(a-b)\text{.}\) Hence the total shaded area is

\begin{equation*} a^2-b^2 = (a-b)(a-b) + 2b(a-b). \end{equation*}

Now each term has a common factor of \((a-b)\text{,}\) hence we may factor it off and simplify by collecting like terms:

\begin{align*} a^2-b^2 \amp = \amp (a-b)(a-b) + 2b(a-b)\\ \amp = \amp (a-b)(a-b+2b)\\ \amp = \amp (a-b)(a+b). \end{align*}

One may also see this factorization geometrically by taking the shaded rectangle on top and moving it to the left hand side of the diagram.

Now we have a single shaded rectangle whose height is \(a-b\) and whose width is \(a+b\text{,}\) hence the total area is \((a-b)(a+b)\text{.}\)