Turning Distance into a Lower Bound
The Triangle Inequality, recalled in the previous tutorial, gives an upper bound: the magnitude of a sum cannot exceed the sum of the magnitudes. Sometimes an upper bound is not what is needed. If two quantities nearly cancel, we may instead want to know how small their sum can be, or how much the magnitude of a number can change when the number is perturbed.
The reverse triangle inequality supplies this kind of information. It was established in Properties of Absolute Value: for real numbers \(u\) and \(v\), the difference between their magnitudes is no greater than the magnitude of their difference. We will use that result, rather than prove it again, to obtain useful lower bounds and magnitude estimates.
Because an absolute value is nonnegative, the reverse triangle inequality immediately implies both \(|u|-|v|\leq|u-v|\) and \(|v|-|u|\leq|u-v|\). Either inequality can be useful as a lower bound after rearranging. The order of the terms matters: the difference on the right is \(u-v\), but its magnitude is unchanged if the order is reversed.
A Lower Bound for the Magnitude of a Sum
To estimate a sum from below, compare one term with the negative of the other. The distance between \(u\) and \(-v\) is \(|u+v|\), so the reverse triangle inequality compares that sum with the difference of the terms’ magnitudes.
Proof. Apply the previously established reverse triangle inequality to the pair \(u\) and \(-v\). It gives
The basic properties of absolute value give \(|-v|=|v|\), and subtraction gives \(u-(-v)=u+v\). Substituting these identities into the inequality proves
Since \(|u|-|v|\leq\bigl||u|-|v|\bigr|\) and \(|v|-|u|\leq\bigl||u|-|v|\bigr|\), both stated one-sided lower bounds follow. \(\square\)
The absolute difference in the theorem is important. If one term has much larger magnitude than the other, the sum cannot be close to zero: a smaller term cannot cancel more than its own magnitude. If the two magnitudes are equal, however, this estimate only says \(|u+v|\geq0\), which is always true. In that case the terms may cancel completely, so a stronger lower bound cannot generally be expected from their magnitudes alone.
Worked Example: A Sum with Unequal Magnitudes
Find a lower bound for \(|-11+4|\) using only the magnitudes of the two terms. The lower-bound theorem gives
The estimate is exact in this case: \(-11+4=-7\), so \(|-11+4|=7\). The smaller term, \(4\), offsets part of the magnitude \(11\), but cannot cancel more than \(4\). The same estimate also works without first calculating the sum.
Worked Example: A Bound That Is Not Exact
Consider \(|9+(-5)|\). The reverse triangle estimate gives
Here the actual sum is \(9-5=4\), so equality holds. To see why this is a natural lower bound, suppose instead that the second term were \(5\). The same estimate would give \(|9+5|\geq|9-5|=4\), which is valid but not exact: \(|14|=14\). A lower bound guarantees that the quantity does not fall below a specified value; it need not predict the exact value.
How Much Can a Magnitude Change?
The reverse triangle inequality also controls changes in absolute value. If a number \(x\) is close to a reference number \(a\), then \(|x|\) must be close to \(|a|\). More precisely, the change in magnitude is at most the distance between \(x\) and \(a\). The next result packages this observation as upper and lower bounds that can be used directly.
Proof. By the reverse triangle inequality applied to \(x\) and \(a\),
From \(|x|-|a|\leq r\), adding \(|a|\) to both sides gives \(|x|\leq|a|+r\). From \(|a|-|x|\leq r\), rearranging gives \(|a|-r\leq|x|\). Also \(|x|\geq0\), by the basic properties of absolute value. Thus \(|x|\) is at least both \(0\) and \(|a|-r\), which means
Combining the lower and upper bounds proves the result. \(\square\)
The maximum with \(0\) handles an important edge case. If \(r>|a|\), then \(|a|-r\) is negative, and that negative number is not a useful lower bound for an absolute value. The fact that \(|x|\geq0\) gives the meaningful lower bound instead. If \(r\leq|a|\), the estimate says that \(x\) cannot move so far from \(a\) that its magnitude drops below \(|a|-r\).
Worked Example: Magnitude Near a Positive Reference Value
Suppose \(|x-6|\leq0.4\). Apply the perturbation theorem with \(a=6\) and \(r=0.4\). Since \(|a|-r=6-0.4=5.6\), the theorem gives
The lower bound is positive, so \(x\) cannot be zero. In fact, the original condition also says that \(x\) lies within \(0.4\) of \(6\); the magnitude estimate records the consequence for \(|x|\) without needing to solve that interval condition. For a numerical check, \(x=5.8\) satisfies \(|5.8-6|=0.2\leq0.4\), and its magnitude \(5.8\) lies between \(5.6\) and \(6.4\).
Using Distance Estimates to Control Expressions
The perturbation theorem is useful when a quantity is known to be close to a simpler reference value. One can estimate the magnitude of the complicated quantity by first estimating its distance from the reference, then adjusting the reference magnitude by at most that distance. This avoids expanding an expression or determining its sign.
For example, suppose \(t\) is close to \(-3\), and \(|t-(-3)|\leq0.2\). The theorem with \(a=-3\) and \(r=0.2\) gives \(2.8\leq|t|\leq3.2\). The negative sign of the reference value does not change its magnitude: \(|-3|=3\). A distance estimate works around positive and negative reference values in exactly the same way.
Worked Example: A Perturbed Negative Number
Suppose \(|y+2|\leq0.7\). Rewrite the expression as \(|y-(-2)|\), so the reference value is \(a=-2\) and the allowed distance is \(r=0.7\). The theorem yields
and therefore
For a direct check, take \(y=-1.5\). Then \(|y+2|=|0.5|=0.5\leq0.7\), while \(|y|=1.5\) satisfies \(1.3\leq1.5\leq2.7\). The estimate also ensures that every \(y\) satisfying the given distance condition is nonzero.
The lower-bound theorem for sums and the perturbation theorem are closely related uses of the same idea. In the first, we compare \(u\) with \(-v\), so their distance is the magnitude of a sum. In the second, we compare \(x\) with a reference value \(a\), so their distance measures how far the input has changed. In each case, the reverse triangle inequality turns a distance into information about magnitudes.
A common pitfall is to assume that a lower bound is an equality, or to use \(|u|-|v|\) as if it were always positive. The safe sum estimate is \(\bigl||u|-|v|\bigr|\leq|u+v|\). If the magnitudes are equal, the estimate gives only zero; if they are unequal, it gives a positive lower bound. Check which situation applies before drawing a conclusion about cancellation.
Check Your Understanding
Use the reverse triangle inequality or the results proved here to answer the following questions.
- What lower bound does the theorem give for \(|14+(-6)|\) using only the magnitudes of the terms?
- If \(|x-(-4)|\leq0.5\), what upper and lower bounds for \(|x|\) follow from the perturbation theorem?
- If \(|u|=|v|\), what lower bound does the sum theorem provide for \(|u+v|\)? Does that bound rule out complete cancellation?
- Suppose \(|z-2|\leq3\). Apply the perturbation theorem, including the maximum with zero, to give a lower and an upper bound for \(|z|\).
- Explain why the reverse triangle inequality can be useful when the sign of a number is unknown.