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Comparing Candidate Values Exactly

Compare candidate values exactly so that radicals, fractions, and parameter-dependent expressions lead to reliable absolute-extremum conclusions.

Intermediate 9 min read

What You'll Learn

  • Compare radical expressions by using sign conditions and equivalent squared inequalities.
  • Order fractions exactly by using common denominators or cross multiplication.
  • Find parameter ranges in which one candidate value exceeds another.
  • Recognize when rounded decimal values cannot establish which candidate is larger.
  • Check that an exact comparison preserves the direction and conditions of the inequality.

Why Exact Comparison Matters

Finding candidate points is only part of an absolute-extrema problem. Once the interior and boundary analysis has produced a finite list of candidates, the largest and smallest function values still have to be identified. Those values may involve fractions, radicals, or a parameter, and their ordering is not always clear by inspection.

A decimal approximation can be useful for forming a conjecture about the ordering. It is not, by itself, a proof. In particular, two different exact values can round to the same decimal, or a small rounding error can reverse the apparent ordering. Exact comparisons settle the issue without relying on displayed precision.

Definition: An exact comparison establishes the order of two values by a logically equivalent equation or inequality, with all sign conditions stated. For a finite list of candidate values, exact comparisons determine which value is largest and which is smallest.

In the absolute-extrema setting, suppose the function is continuous on a compact region and the candidate analysis has produced all points where an absolute maximum or minimum can occur. The extreme values must be among the function values at those candidates. The task in this tutorial is to compare that list, not to find the candidates.

A Reliable Comparison Strategy

For two candidate values \(A\) and \(B\), the basic question is whether \(A-B\) is positive, negative, or zero. The transformation used to determine its sign must preserve the comparison. A useful procedure is:

1
Record the exact values.
Keep radicals and fractions in exact form rather than replacing them with rounded decimals.
2
Check signs and denominators.
Before multiplying an inequality or squaring both sides, establish the conditions that make the operation valid.
3
Use an equivalent comparison.
Subtract, use a common denominator, or square nonnegative quantities as appropriate.
4
State the ordering.
Use the proved comparisons to identify the largest and smallest candidate values.

For fractions with positive denominators, cross multiplication is safe: if \(b>0\) and \(d>0\), then \(\frac{a}{b}<\frac{c}{d}\) exactly when \(ad<bc\). If a denominator might be negative, multiplying by it can reverse the inequality, so its sign must be checked first.

For radicals, squaring can turn a comparison into an easier one, but it preserves the comparison directly only when both sides are nonnegative. If \(A\geq 0\) and \(B\geq 0\), then \(A<B\) exactly when \(A^2<B^2\). Without those sign conditions, the conclusion need not hold.

Worked Example: Ordering Values with Radicals

Worked Example: Ordering Values with Radicals

Suppose three candidate points \(P,Q,R\) give the values \(3\), \(1+\sqrt{5}\), and \(2\sqrt{3}\), respectively. Determine their order exactly and identify the largest and smallest.

First compare \(3\) and \(1+\sqrt{5}\). Since \(5>4\) and both square roots are nonnegative, \(\sqrt{5}>\sqrt{4}=2\). Therefore,

$$ 3<1+\sqrt{5}. $$

Next compare \(1+\sqrt{5}\) and \(2\sqrt{3}\). Both are positive, so it is valid to compare their squares:

$$ (1+\sqrt{5})^2=1+2\sqrt{5}+5=6+2\sqrt{5}, \qquad (2\sqrt{3})^2=4\cdot 3=12. $$

Because \(5<9\), we have \(\sqrt{5}<3\). Thus \(2\sqrt{5}<6\), and so \(6+2\sqrt{5}<12\). The original positive values therefore satisfy \(1+\sqrt{5}<2\sqrt{3}\). The complete order is

$$ 3<1+\sqrt{5}<2\sqrt{3}. $$

Hence \(P\) gives the smallest candidate value and \(R\) gives the largest. The comparisons check out: the first follows from \(\sqrt{5}>2\), and the second follows from \((1+\sqrt{5})^2< (2\sqrt{3})^2\) with both values positive.

Why Comparing Squares Can Work

The preceding example illustrates a general justification. For nonnegative real numbers \(A\) and \(B\),

$$ A^2-B^2=(A-B)(A+B). $$

If \(A\) and \(B\) are not both zero, then \(A+B>0\). Multiplication by the positive number \(A+B\) preserves the sign, so \(A^2-B^2\) and \(A-B\) have the same sign. Consequently, \(A^2<B^2\) if and only if \(A<B\). If both are zero, both comparisons are equalities, so the equivalence still holds.

This proof explains both the usefulness and the limitation of squaring. The nonnegativity assumption is essential. For instance, \(-3<2\), but \((-3)^2>2^2\). Squaring those two sides has not preserved their order because the left side was negative.

Comparison rule: To infer \(A<B\) from \(A^2<B^2\), first verify \(A\geq0\) and \(B\geq0\). If either sign is unknown, compare the original expressions by another method or establish their signs before squaring.

Worked Example: Fractions and a Radical

Worked Example: Fractions and a Radical

Suppose the candidate values are \(\sqrt{3}\), \(\frac{7}{4}\), and \(\frac{9}{5}\). Find their exact order.

All three values are positive. To compare \(\sqrt{3}\) with \(\frac{7}{4}\), square both nonnegative values:

$$ (\sqrt{3})^2=3=\frac{48}{16}, \qquad \left(\frac{7}{4}\right)^2=\frac{49}{16}. $$

Since \(\frac{48}{16}<\frac{49}{16}\), it follows that \(\sqrt{3}<\frac{7}{4}\). Now compare the two fractions. Their denominators are positive, so cross multiplication gives

$$ \frac{7}{4}<\frac{9}{5} \quad\Longleftrightarrow\quad 7\cdot5<9\cdot4 \quad\Longleftrightarrow\quad 35<36. $$

Therefore the full order is

$$ \sqrt{3}<\frac{7}{4}<\frac{9}{5}. $$

The smallest candidate value is \(\sqrt{3}\), and the largest is \(\frac{9}{5}\). The exact checks are \(48<49\) for the squared comparison and \(35<36\) for the fraction comparison.

Comparisons That Depend on a Parameter

Sometimes the candidate values contain a parameter, and their order can change as that parameter varies. The result should then specify the parameter ranges, including any value where the candidates tie. A key step is to turn the comparison into a sign question, while keeping track of the assumptions that make each transformation valid.

Worked Example: Finding a Parameter Threshold

For \(0\leq t\leq1\), suppose two candidates have values \(A(t)=\sqrt{t+2}\) and \(B(t)=t+1\). Determine which is larger for each parameter value.

Since \(0\leq t\leq1\), both \(t+2\) and \(t+1\) are positive, so \(A(t)\) and \(B(t)\) are nonnegative. Squaring therefore preserves their order. We obtain

$$ \begin{aligned} A(t)>B(t) &\Longleftrightarrow t+2>(t+1)^2\\ &\Longleftrightarrow t+2>t^2+2t+1\\ &\Longleftrightarrow t^2+t-1<0. \end{aligned} $$

The positive root of \(t^2+t-1=0\) is \(\alpha=\frac{\sqrt{5}-1}{2}\). Indeed, \(\alpha^2+\alpha-1=0\). The polynomial factors as

$$ t^2+t-1=(t-\alpha)(t+\alpha+1). $$

For \(t\geq0\), the second factor \(t+\alpha+1\) is positive. Thus the expression is negative when \(t<\alpha\), zero when \(t=\alpha\), and positive when \(t>\alpha\). Within the given interval,

$$ \begin{cases} A(t)>B(t),&0\leq t<\alpha,\\ A(t)=B(t),&t=\alpha,\\ A(t)<B(t),&\alpha<t\leq1. \end{cases} $$

For a direct check on either side of the threshold, take \(t=\frac12\). Then \(A(t)^2=\frac52=\frac{10}{4}\), while \(B(t)^2=\frac94\), so \(A(t)>B(t)\). At \(t=\frac34\), \(A(t)^2=\frac{11}{4}=\frac{44}{16}\), while \(B(t)^2=\frac{49}{16}\), so \(A(t)<B(t)\). At \(t=\alpha\), the defining equation \(\alpha^2+\alpha-1=0\) gives \(\alpha+2=(\alpha+1)^2\); both sides correspond to nonnegative values, confirming \(A(\alpha)=B(\alpha)\).

When Decimal Approximations Are Not Enough

A decimal can conceal a distinction that matters. If two values both display as \(1.414\) to three decimal places, that display alone does not show whether they are equal or which is larger. More digits may suggest the answer, but an exact inequality is what establishes it.

Worked Example: Two Values That Round the Same Way

Compare \(\sqrt{2}\) and \(1.414=\frac{707}{500}\). To compare them exactly, note that both are positive and calculate

$$ \left(\frac{707}{500}\right)^2 =\frac{707^2}{500^2} =\frac{499849}{250000} < \frac{500000}{250000} =2 =(\sqrt{2})^2. $$

Since both values are nonnegative, the squared comparison proves \(\frac{707}{500}<\sqrt{2}\). Numerically, \(\sqrt{2}\) is approximately \(1.414213\), which also rounds to \(1.414\) to three decimal places. The rounded displays match, but the exact values do not. The comparison is checked by squaring: \(499849<500000\).

This distinction matters when candidate values are close, especially if they determine which point gives an absolute maximum or minimum. An approximation that does not resolve the ordering should be treated as inconclusive. Use it to guide the choice of an exact comparison, then make the final decision using that comparison.

A Common Pitfall: Changing an Inequality Without Checking Signs

A frequent error is to square both sides of an inequality without checking whether the values are nonnegative. For example, starting from \(-3<2\) and squaring gives \(9>4\), not \(9<4\). The correct conclusion from the squared values cannot be read as though squaring preserved the original order.

For radical comparisons, first verify that each side is nonnegative. For fraction comparisons, verify that denominators are positive before cross multiplying; a negative multiplier reverses the inequality. For parameter comparisons, do not omit the parameter range needed to justify those steps. These checks are part of the comparison, not optional details.

Takeaway: Exact comparisons depend on valid transformations. State the sign conditions, preserve the direction of inequalities, and include equality cases when parameters are involved.

Check Your Understanding

For each question, give an exact argument rather than relying only on decimal approximations.

  1. Which is larger: \(\sqrt{6}\) or \(\frac{5}{2}\)? What sign condition allows you to compare their squares?
  2. Order \(\frac{11}{6}\) and \(\frac{7}{4}\) by cross multiplication.
  3. For \(t\geq0\), compare \(\sqrt{t+1}\) and \(1\). When are they equal?
  4. Why does \(-4<3\) not imply \((-4)^2<3^2\)?
  5. Two candidate values both round to \(2.718\) to three decimal places. Does that establish that they are equal? Explain.

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