Equivalent modifications of inequalities

When modifying inequalities, we use equivalent modifications, which are characterized by the fact that they do not change the validity of the inequality. The point of equivalent adjustments is to get the inequality into some simpler form from which we can already calculate the result of the inequality.

What is an inequality

We are given two functions f(x) and g(x). We then understand an inequality to be an expression in one of these forms:

$$\begin{eqnarray} f(x) < g(x),&\quad& f(x) > g(x)\\ f(x)\le g(x),&\quad& f(x)\ge g(x) \end{eqnarray}$$

The solutions of an inequality are all x of the (usually) set of real numbers for which the inequality is satisfied. Example inequality: 6x<12. Here, f(x) = 6x and g(x) = 12. 6x is the left-hand side of the inequality, 12 is the right-hand side. The solutions to this inequality are all x, which are less than two: x<2. Thus, the interval (−∞, 2).

Changing sides with a change of sign

For the equations, we could easily swap the right side for the left side. We cannot do this for inequalities. For example, we have an inequality x<5, so it cannot also be 5 < x. In the first case x is less than five, in the second case x is greater than five. However, if we change the sign of the inequality at the same time as swapping the side, it will already be an equivalent adjustment.

Thus, if we have the inequality f(x) < g(x), then we can also write g(x) > f(x). We have changed the sides, but at the same time we have written > instead of <. Similarly for less than or equal to: We can modify the inequality f(x)≤ g(x) to g(x)≥ f(x).

Adding an expression

We can add a number or function to an inequality that is defined on the same domain as the domain we are solving the inequality on. Example: we have the inequality x>0. For example, we can add the number three to this inequality. We get the inequality x + 3>3. This makes sense.

Similarly, we can add the whole function. If we have the inequality x > −3x + 7, we can add 3x to the inequality and get:

$$\begin{eqnarray} x& > &-3x+7\quad/+3x\\ x+3x& > &-3x+3x+7\\ 4x& > &7 \end{eqnarray}$$

Of course, we can subtract the expression that way - or we can add a negative number.

Multiplying an inequality by a positive expression

We can't just multiply an inequality by an expression like we could do with equations. The rule is that as long as we multiply by an expression that is always positive, it's fine. But if we multiply by a negative expression, we have to change the sign (see the next chapter). Of course, as with equations, we must not multiply by a zero expression!

What does a positive expression always mean? The simplest expression that is always positive is a positive number. So if we have an inequality x>2, we can multiply it by ten, which is a positive number, and the validity of the inequality will not change. We get the inequality 10x>20. Of course, we can also divide by a positive number, so we can get the inequality 2x>4 from this modified inequality by dividing by five.

But we can't just multiply the inequality by the variable x, because in general the variable x can take on both positive and negative values, and this is not allowed. But there are situations where x will always be positive. For example, if we are calculating some geometric example, it is possible that the variable x will represent the length of some line segment. The length cannot be negative (at worst zero), so if we know that x will equal the length of some line segment, we can multiply the variable x by the inequality.

The next always non-negative expression is the square of the square. If you square a number, it will be positive (or zero). So if we have an inequality

$$\frac{1}{x^2}>2,$$

we can multiply it by the expression x². We get: 1>2x². Provided, of course, that x is not equal to zero. There are multiple functions that are always positive. Such Goniometric functions sine and cosine have as their domain of values the set <−1,1>. This is not always positive. But if we add two to them, we get the range of values of <1, 3>, which is already a positive interval. We could multiply an inequality by such a function. So if we had an inequality

$$\frac{x+3}{\cos(x)+3}\le 3x-5,$$

we can multiply the whole thing by the expression cos(x)+3, because that's always a positive function. We get:

$$x+3\le(3x-5)(\cos(x)+3).$$

Multiplying an inequality by an always negative expression

With negative numbers and expressions, the situation is more complicated. See the example: 1<2. This inequality certainly holds. What if we multiply the whole inequality by minus one? We get −1<−2. Is this inequality valid? No, minus two is less than minus one. Obviously, multiplying by a negative number is not equivalent to modifying the inequality.

However, if when multiplying by a negative number in an inequality we also change the sign of the inequality, it is already an equivalent adjustment. If we multiply the inequality 1<2 by −1 and at the same time write > instead of <, we make an equivalent adjustment and get the inequality −1>−2.

Multiplication by negative numbers is easy, the situation is worse for multiplying expressions (functions). Why can't we multiply the inequality by the variable x? Because if the variable took positive values, it would not change sign. If it took negative values, we would have to change the sign. In general, both cases can occur, if we don't have x constrained in any way, the variable can be both negative and positive. Example:

$$\frac{1}{x}>1$$

The wrong approach would be to multiply the inequality by the variable x. This is because we would get 1>x, so the solution is all x that are less than one (still with the constraint that x is different from zero, of course). But what if we substitute the number −1 into the original inequality after x? We get:

$$\begin{eqnarray} \frac{1}{-1}&>&1\\ -1&>&1 \end{eqnarray}$$

which is certainly not true. Therefore, we can't just multiply the inequality by the variable x.

However, just as we had expressions that are always positive in the previous chapter, we can have expressions that are always negative. The easiest way to get them is to put a minus sign in front of an always positive expression. In the previous chapter, we noted that the function cos(x)+3 is always positive. If we put a minus sign in front of the whole expression, we get an always negative function: −(cos(x)+3).

Multiplying and subtracting an inequality by a natural number

The amplification of an inequality is generally not an equivalent treatment, as with equations. Consider the following inequality: x>−2. What if we factor it squared? We get: x²>4. Is this an equivalent inequality? What happens if we substitute a zero after x into both equations? In the first case, we have 0>−2, which is true. In the second case, we have 0>4, which is not true. So we haven't made an equivalent adjustment.

When can we multiply an inequality by a natural number? If both sides are non-negative (positive or zero). Then it will be no problem. Example:

$$\begin{eqnarray} 5&>&2\quad/^2\\ 25&>&4 \end{eqnarray}$$

Another example might be an absolute value that returns a non-negative number. So we can safely factor this inequality into the other:

$$\begin{eqnarray} |x|&<&3\quad/^2\\ |x|^2&<&9 \end{eqnarray}$$

A square root behaves exactly the same as a power. If both sides are non-negative, we can square the equation with the natural square root. So we can nicely square root the previous inequality back to its original form:

$$\begin{eqnarray} |x|^2&<&9\quad/\sqrt{ }\\ |x|&<&3 \end{eqnarray}$$

Removing unnecessary expressions

Often we solve an inequality that is of the form f(x)>0. In this case, the function f is complex and could be simplified by dropping some expressions. Take a look at the following inequality: 2x>0. This inequality obviously has a solution when x>0. If x is negative, so is 2x. If we multiply a negative number by two, the number remains negative. A two does not have the power to change the sign of the following expression after multiplication.

We can still try to solve the inequalities 5x>0, 157x>0, and π x>0. In all cases, the solution will be the same, x>0. A positive coefficient before x will not change the validity of the inequality.

Formally, we then get rid of these coefficients by dividing the inequality by the given coefficient, as follows:

$$\begin{eqnarray} 2x&>&0\quad/:2\\ x&>&0 \end{eqnarray}$$

Note that in order to remove the positive coefficient, let's label it a, it must necessarily multiply the entire left-hand side. We cannot remove the two from this inequality: 2x + 7>0, because the two does not multiply the whole left-hand side. But we can divide the whole equation by two and get the inequality x + 3,5>0.

Similarly, this example:

$$x^2\cdot(x-3)>0$$

It may look complicated, but the expression x² is always non-negative, so it doesn't change the sign of the following parenthesis. So we can remove this expression from the inequality - divide the inequality by x² and we get:

$$x-3>0$$

We just need to add the condition that x²≠0, because we can't divide by zero. However, x² is only zero if x is zero and for x = 0 the inequality is not satisfied. This is because we get (x = 0): 0(0 − 3)>0 and 0>0 is not true.

This is not an extra method, it's just a handy application of the previous point about multiplying and dividing by a positive expression.