Bisection without continuity: a small bracket, no root

Bisection without continuity can produce an interval that keeps getting smaller without enclosing any root at all. The endpoint signs can remain opposite. Every midpoint update can follow the stated rule correctly. The missing fact is whether the original function ever takes the value zero. This exercise makes that distinction explicit with a function whose only outputs are minus one and plus one. Its intervals converge to a perfectly definite location, but that location is a jump boundary. A second, continuous example shows how the same sort of interval calculation behaves when a root really is present.

Use exact real arithmetic for the argument. There is no measurement error, rounding error or software timing claim to diagnose. The interval endpoints are ordinary rational numbers, and the first example is defined at every real input. We will retain a negative left endpoint and a positive right endpoint throughout. Those conditions describe the signs of two evaluations; they do not, by themselves, describe every evaluation between them. Keeping that distinction visible lets us explain the failure without inventing a broken update or changing the function midway through the calculation.

Specify the value at the jump

Define f(x) to equal −1 whenever x is strictly less than zero, and +1 whenever x is greater than or equal to zero. In particular, f(0) = +1. That final assignment matters: leaving the value at zero unspecified would create a different problem about an incomplete definition. Here every input belongs to exactly one branch, so every evaluation is available. On the negative half of the number line the answer is always −1. At zero and everywhere to its right the answer is always +1.

A root would be a real number r satisfying f(r) = 0. Neither branch returns zero. Therefore this function has no root anywhere, including inside the starting interval [−1, 1]. This proof does not depend on how many midpoint evaluations we perform. It follows directly from the exhaustive definition of the function. Before running the interval calculation, we already know what a claimed root would have to contradict: it would require one of the two declared outputs to equal zero.

The starting endpoints nevertheless pass an opposite-sign check: f(−1) = −1 and f(1) = +1. A graph represents the left branch by a horizontal segment at height −1 ending with an open circle at x = 0. The right branch begins with a filled circle at height +1 and x = 0. There is no vertical function segment joining those circles. Drawing such a segment would falsely give the function intermediate values that its definition never assigns, including the zero we are trying to investigate.

Run the correct midpoint updates

Let L and R denote the current left and right endpoints. Set m = (L + R)/2 and evaluate the original f at m. If that evaluation is negative, replace L by m. If it is positive, replace R by m. An exact zero would settle the root question immediately, but our declared f never produces one. The update keeps the negative endpoint on the left and the positive endpoint on the right. Each retained interval is closed and lies inside the preceding interval.

From [−1, 1], the first midpoint is zero. Because f(0) = +1, the positive-endpoint update sets R to zero and leaves L at −1. The first retained interval is [−1, 0], of width one. Nothing has gone wrong in that step. Zero has the same positive sign as the old right endpoint, so retaining it as the new right endpoint follows the stated rule exactly. Calling the endpoint update incorrect would hide the actual missing assumption.

The next midpoint is −1/2, where f is negative. Replace the left endpoint to obtain [−1/2, 0], with width 1/2. The following midpoint is −1/4, giving [−1/4, 0], with width 1/4. Then come [−1/8, 0] and [−1/16, 0]. Throughout this trace the right endpoint stays at zero. Each newly evaluated midpoint is negative, so each later step moves only the left endpoint. The endpoint values remain −1 and +1 in every displayed interval.

Index the retained intervals by n, where n = 1 means one completed update from the original [−1, 1]. Their exact formula is L_n = −2^(1−n) and R_n = 0. The width R_n − L_n is 2^(1−n). The first update establishes the formula at n = 1. If it holds at some later step, the midpoint is L_n/2, which is still negative. Replacing L_n by that midpoint divides its magnitude by two and leaves R_n unchanged, establishing the formula for the next step.

This indexing also prevents a misleading numerical report. After one update the width is one, not one half, because the original width was two. After ten completed updates it is 1/512. Either starting width could be used in a different example, but the count and formula must refer to the same starting interval. These bookkeeping details can be checked independently of root existence. Getting every interval width right still leaves the no-root proof from the function definition completely intact.

Read the limit and the residual separately

The left endpoints approach zero from below, while every right endpoint equals zero. The interval widths approach zero as well. Thus the intervals identify a unique limiting coordinate: zero. That is a valid statement about the endpoints. At that coordinate, however, the original function returns +1. The limiting coordinate is therefore not a solution of f(x) = 0. A definite limiting location and a root are different mathematical claims, and this example satisfies only the first.

We can also inspect a residual directly. For a proposed input x, define the absolute residual as |f(x)|, since the target function value is zero. In this example the residual equals one for every real x. It equals one at both endpoints, at every evaluated midpoint and at the midpoint of every retained bracket. After ten steps, after twenty steps, or after any other finite number, the residual has not decreased. The shrinking horizontal width measures a coordinate interval; the residual measures the output's distance from zero.

For example, the interval after four updates is [−1/8, 0]. Its midpoint is −1/16 and its width is 1/8, yet f(−1/16) = −1. The absolute residual is one. If a report states only that the width is below a chosen horizontal tolerance, it has reported that interval fact correctly. It has not thereby established a small output residual or the existence of a zero. Reporting all three quantities would expose the difference immediately: a small width, a residual of one and an explicit no-root definition.

The jump explains why the endpoint signs never force a zero here. Inputs arbitrarily close to zero on the left still return −1, whereas the value at zero is +1. Approaching the same coordinate does not make those function values approach the value assigned there. The calculation does not require a theorem about all possible discontinuous functions: our two-branch definition already proves the needed counterexample. One such fully specified function is enough to refute the claim that opposite signs and shrinking width alone always certify a root.

Keep a continuous control beside the counterexample

Now define g(x) = x − 1/3 on [0, 1]. Its exact root is 1/3, because substituting that input gives zero. The endpoint values are −1/3 and +2/3. Use the same midpoint rule, retaining a negative left endpoint and a positive right endpoint. The first midpoint is 1/2, so the first retained bracket is [0, 1/2]. The next midpoint is 1/4, giving [1/4, 1/2]. Evaluating 3/8 next gives [1/4, 3/8]. Every one of those brackets contains 1/3.

The reason root retention works can be proved from this particular formula. The sign of g(x) is negative exactly when x is below 1/3 and positive exactly when x is above 1/3. If a midpoint is negative, the root remains to its right; if it is positive, the root remains to its left. Replacing the endpoint with the matching sign therefore retains the known root in either case. This proof concerns the linear control as defined, without pretending that the step function shares its behavior.

After n updates the control's bracket width is 2^(−n), since its original interval has width one. Let c_n be the midpoint of the retained bracket. Because 1/3 lies inside that bracket, its distance from c_n is at most half the bracket width. For our linear function, |g(c_n)| equals |c_n − 1/3| exactly. Consequently its midpoint residual is at most 2^(−n−1). Here we have an explicit connection between horizontal interval width and output residual, supplied by the particular linear formula.

No midpoint in this control equals 1/3 exactly. Starting from zero and one and repeatedly taking midpoints produces fractions whose reduced denominators are powers of two. The reduced denominator of 1/3 is three. Exact equality would require those reduced forms to represent the same rational number, which they cannot. The brackets can therefore contract indefinitely around a root without ever evaluating the root exactly. That outcome differs from the step-function example, where indefinite contraction occurs even though no root exists at all.

Two decorative accounts stay outside the proof

The Studio 55D McGinn decorative vase account shows a rounded vessel with a short circular neck and a striated surface. Its gallery includes detail crops, room staging and a dimension graphic. Those pictures do not supply a function for this exercise, and the catalog name does not authenticate a maker, material or physical measurement.

The twisted decorative vase account shows a tall twisting vessel with an angular mouth and base. Black and white comparison images appear, while the selected variant is Black. Flowers and tableware are staging, not an established bundle. This form differs from the rounded vessel, but neither silhouette proves continuity, a residual bound or a root.

Nearby problems require their own reasoning

The first-true search guide studies an integer threshold and an endpoint update that discards a passing midpoint. Its repair retains that candidate. The updates in our real-interval counterexample already retain the correct sign endpoints; changing an integer boundary convention would not create a zero in f.

The extraneous solution after squaring guide checks transformed-equation candidates in the original radical equation. Our example never transforms its equation or switches its evaluation rule. Every residual is computed from the same original two-branch f, whose output set excludes zero from the beginning.

Describe the result the calculation earned

For the step function, an accurate result statement is: the correct sign-preserving midpoint updates generate brackets of width 2^(1−n) approaching zero, their common limiting coordinate is zero, and the original function has no root. For the continuous control, the brackets retain the known root 1/3, their widths are 2^(−n), and the midpoint residual is bounded by half that width. These statements preserve what each calculation actually demonstrates. A shared pattern of interval contraction does not erase the difference between the two functions' values.

If a later purchase uses cryptocurrency

The mathematical examples do not establish anything about a purchase or a payment's reversibility. If you later consider one of the linked listings, check the exact selected item and its current terms at the time of ordering. If that checkout offers cryptocurrency and you choose it, read the particular payment request before sending funds. Confirm that the requested asset and network match the ones you intend to use, and check the destination and amount against the order's instructions. A familiar currency symbol does not resolve an asset or network mismatch, and the signs in our function exercise have no bearing on a merchant's payment instructions.

Keep the order identifier and any payment reference supplied by the checkout. If its status remains uncertain after an attempted payment, reconcile that attempt through the order's stated support process before making another payment. A pending screen alone is not evidence that an earlier transfer never happened. Review any stated fees, confirmation requirements and refund procedure before deciding whether to proceed. This article does not promise that a cryptocurrency option is available for either item, that a transfer will settle at a particular speed, or that a refund will use the same asset or network. Those questions depend on the actual order and terms presented to you.