Proofs Involving Absolute Values

Absolute Values as a Proof Tool

An absolute value measures distance from zero, but its main role in proofs is often to control a quantity without first determining its sign. The previous tutorial used the Well-Ordering Principle and induction to reason about integers. Here we turn to a different kind of proof technique: use the definition of absolute value, and choose carefully whether to split into sign cases or work with inequalities.

The earlier tutorials on reverse triangle inequalities and absolute-value inequalities established useful ways to interpret bounds such as $|x|<r$. We will build on those results. Our central new result is the triangle inequality, which bounds the absolute value of a sum using the sizes of its terms.

Definition. For $x\in\mathbb{R}$, the absolute value of $x$ is $$ |x|=\begin{cases} x,&x\geq0,\\ -x,&x<0. \end{cases} $$ In particular, $|x|\geq0$, $|x|^2=x^2$, and $|x|=0$ exactly when $x=0$.

The definition gives a direct method for proofs: if the sign of an expression is known, remove the absolute value using the appropriate case. If its sign is unknown, splitting into cases may be cumbersome; inequalities can often handle all signs at once.

Basic Algebra and the Triangle Inequality

Two elementary identities are useful throughout. For real $x$ and $y$, $|xy|=|x||y|$, and for real $c$, $|cx|=|c||x|$. To verify the product identity, both sides are nonnegative and

$$ (|xy|)^2=x^2y^2=(|x||y|)^2. $$

Two nonnegative real numbers with equal squares are equal, so $|xy|=|x||y|$. Taking $y=c$ gives the scaling identity. We will also use $|t|\geq t$, which follows directly from the definition: if $t\geq0$, then $|t|=t$, and if $t<0$, then $|t|=-t>t$.

Theorem (Triangle Inequality). For all $x,y\in\mathbb{R}$, $$ |x+y|\leq |x|+|y|. $$

Proof. Set $A=|x|+|y|$ and $B=|x+y|$. Both $A$ and $B$ are nonnegative. Using $|x||y|=|xy|$ and the identity $|t|^2=t^2$, we calculate

$$ A^2-B^2 =x^2+2|xy|+y^2-(x+y)^2 =2(|xy|-xy). $$

Since $|xy|\geq xy$, the final expression is nonnegative. Thus $A^2\geq B^2$. Because $A,B\geq0$, this implies $A\geq B$: if $A<B$, then $B^2-A^2=(B-A)(B+A)>0$, a contradiction. Therefore $|x+y|\leq|x|+|y|$. $\square$

This proof illustrates an important way to use squares: first confirm that both quantities being compared are nonnegative. Squaring an inequality between quantities of unknown sign is not generally valid, but comparing squares of nonnegative quantities preserves their order.

Theorem (Equality in the Triangle Inequality). For $x,y\in\mathbb{R}$, $$ |x+y|=|x|+|y| \quad\Longleftrightarrow\quad xy\geq0. $$

Proof. With $A$ and $B$ as above, the calculation in the triangle inequality proof gives

$$ A^2-B^2=2(|xy|-xy). $$

If $xy\geq0$, then $|xy|=xy$, so $A^2=B^2$. Since both numbers are nonnegative, $A=B$. Conversely, if $A=B$, then $A^2-B^2=0$, so $|xy|=xy$. By the definition of absolute value, this holds exactly when $xy\geq0$. This proves both directions. $\square$

The equality condition has a simple interpretation: the two terms have the same sign, or at least one is zero. When they have opposite signs, adding them partly cancels their magnitudes, so the inequality is strict.

Worked Examples

Worked Example: Bounding a Linear Combination

For real numbers $a$ and $b$, prove that

|3a-2b|\leq3|a|+2|b|.

Write the expression as a sum, $3a+(-2b)$, and apply the triangle inequality. The scaling identity gives $|3a|=3|a|$ and $|-2b|=2|b|$, since $|3|=3$ and $|-2|=2$. Hence

|3a-2b| =|3a+(-2b)| \leq |3a|+|-2b| =3|a|+2|b|.

No assumptions about the signs of $a$ and $b$ are needed. A sign-case proof would have to consider several possibilities; the triangle inequality handles them together.

Worked Example: Using a Known Distance to Bound Another Expression

Suppose $|x-4|\leq1$. Prove that $|2x-7|\leq3$, without first solving for $x$.

Rewrite the target expression so that the known quantity $x-4$ appears:

$$ 2x-7=2(x-4)+1. $$

The identity is verified by expanding the right side: $2(x-4)+1=2x-8+1=2x-7$. Now apply the triangle inequality and scaling identity:

|2x-7| =|2(x-4)+1| \leq2|x-4|+1 \leq2\cdot1+1 =3.

The final inequality uses the hypothesis $|x-4|\leq1$. This style of argument is useful when a target expression can be written as a multiple of a controlled error plus a fixed term.

Worked Example: A Strict Bound for a Difference

Suppose $|x|<2$ and $|y|\leq3$. Prove that $|x-y|<5$.

Apply the triangle inequality to $x+(-y)$, then use $|-y|=|y|$:

|x-y|\leq|x|+|-y|=|x|+|y|<2+3=5.

The conclusion is strict because one of the two bounds being added is strict: $|x|<2$. The other is non-strict, $|y|\leq3$. Replacing both hypotheses by non-strict bounds would give only $|x-y|\leq5$, not necessarily a strict inequality.

Worked Example: A Lower Bound from the Triangle Inequality

Prove that, for every real $x$,

|x+1|+|x-1|\geq2.

Apply the triangle inequality to the two terms $x+1$ and $-(x-1)$. Their sum is

$$ (x+1)-(x-1)=x+1-x+1=2. $$

Therefore

2=|2| =|(x+1)-(x-1)| \leq|x+1|+|-(x-1)| =|x+1|+|x-1|.

The equality condition can also be identified. The two terms in this application are $u=x+1$ and $v=-(x-1)=1-x$. Equality holds exactly when $uv\geq0$. Their product is

$$ (x+1)(1-x)=1-x^2. $$

Thus equality holds exactly when $1-x^2\geq0$, or equivalently $|x|\leq1$. For example, if $x=0$, the left side is $|1|+|-1|=2$. If $x=2$, it is $|3|+|1|=4$, so the inequality is strict.

Choosing a Proof Strategy

There are two common approaches to an absolute-value proof. A sign-case argument is direct when the relevant sign is already known or when there are only a few simple cases. The triangle inequality is usually more efficient when the expression is a sum and the signs of its terms are unknown. For instance, to prove a bound on $|3a-2b|$, there is no need to determine whether $a$, $b$, or the whole expression is positive.

The Reverse Triangle Inequality, proved in the earlier tutorial “Reverse Triangle Inequality,” supplies a complementary kind of bound. The triangle inequality gives an upper bound on the magnitude of a sum; the reverse triangle inequality can give a lower bound when one term is compared with another. These results should be chosen according to the direction of the desired estimate.

A frequent error is to remove absolute-value bars without checking signs. For example, $|x-y|=x-y$ is valid only if $x-y\geq0$. Another is to square both sides of an inequality before confirming the relevant signs. In the triangle inequality proof, nonnegativity was established first, so comparison of squares was legitimate. A third error is to lose strictness: from $|x|<2$ and $|y|\leq3$, the sum is strictly less than $5$, but from $|x|\leq2$ and $|y|\leq3$, that conclusion does not follow.

Key takeaway. Use the definition of absolute value when signs are known, and use the triangle inequality when they are not. Before squaring, verify nonnegativity; before claiming equality or strictness, check the signs and hypotheses that determine it.

Check Your Understanding

Use the definitions and proof techniques in this tutorial to answer the following questions.

Why is it valid to compare $A^2$ and $B^2$ in the proof of the triangle inequality only after establishing $A\geq0$ and $B\geq0$?
For which real $x,y$ does equality hold in $|x+y|\leq|x|+|y|$?
Use the triangle inequality to prove $|4u+v|\leq4|u|+|v|$ for all real $u,v$.
If $|s|<1$ and $|t|\leq4$, what strict upper bound follows for $|s+t|$?
In the last worked example, why can the triangle inequality be applied to $x+1$ and $-(x-1)$ to obtain a lower bound?