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Consider the original worked example $f(x)=x^3+x-1$ on $[0,1]$. First, existence: a polynomial function is a continuous function, and $f(0)=-1<0<1=f(1)$. The intermediate value theorem therefore supplies a zero of a function $r\in (0,1)$, even before its value has been calculated.

Second, uniqueness is a separate argument. The derivative is $f^{\prime }(x)=3x^2+1>0$. The derivative criterion for monotonicity, obtained from the mean value theorem, makes $f$ a strictly increasing function. Two distinct zeros of a function would contradict that strict increase. Thus this root bracket contains exactly one zero of a function; indeed the same derivative argument proves uniqueness on the entire real line.

Third, localization uses the existing bisection method. Set $[a_0,b_0]=[0,1]$; after $k$ completed halvings, retain $[a_k,b_k]$ with opposite endpoint signs. Each update tests the old midpoint $m_{k-1}=(a_{k-1}+b_{k-1})/2$. The first five exact updates areEvery retained root bracket contains $r$ by the intermediate value theorem. After ten halvings, exact arithmetic givesHence $0.681640625<r<0.6826171875$. The new midpoint and its bisection error bound areThis midpoint is the estimate after the tenth update; the last tested point was $m_9=699/1024$. In general the retained width is $2^{-k}$, so the bisection error bound is $2^{-(k+1)}$. Nine halvings already suffice for an error at most $10^{-3}$, since $2^{-10}=0.0009765625<10^{-3}$. This guarantee comes from the width of the root bracket, independently of any displayed decimal approximation to the unknown zero of a function. Endpoint signs must be established reliably if evaluations use floating-point arithmetic.

For coefficients with sum one, the defining identity follows from linearity:The translation appears once on both sides precisely because ${\sum }_i{\lambda }_i=1$. In particular, midpoints map to midpoints, and a parameterized line segment $(1-t)P+tQ$, $0\le t\le 1$, maps to $(1-t)T(P)+tT(Q)$. A whole line has the same formula with $t\in \mathbb{R}$; its image is a line if $M(Q-P)\ne 0$, and a single point otherwise.

In exact arithmetic, after $k$ completed halvings the retained root bracket has width $W_k=b_k-a_k$ and midpoint $m_k$, whereThe absolute value bound holds for every zero of a function $x_{\ast }$ in the current root bracket, including the one approached by continued bisection method updates. A zero of a function discarded with an earlier half need not satisfy later bounds. Before update $k\ge 1$, the evaluated candidate is $m_{k-1}$, whose corresponding guarantee is $(b-a)/2^k$. After update $k$, the new midpoint $m_k$ has the smaller guarantee $(b-a)/2^{k+1}$, even without evaluating $f(m_k)$.

The guaranteed error halves at every step; the actual absolute value error need not decrease at every step or have successive ratio $1/2$. In the square root example it increases from $|m_6-\sqrt{2}|\approx 0.000151062$ to $|m_7-\sqrt{2}|\approx 0.003755188$, while both errors remain below their guarantees. An exact midpoint zero of a function instead gives zero error and immediate termination. For this example $\sqrt{2}$ is an irrational number, so no finite exact halving can produce it as a midpoint.

For a brief proof outline, each retained half has half the previous width and contains a zero of a function by the intermediate value theorem. Every point in the retained root bracket is at most half its width from its midpoint. If the process does not stop at a zero of a function, the nested interval theorem gives one common point as the widths tend to zero; continuity and the opposite endpoint signs force its function value to be zero. Uniqueness in the initial root bracket is unnecessary.

The bisection method finds a zero of a function by repeatedly halving a root bracket. Let $f$ be a real-valued continuous function on the closed interval $[a,b]$, with $a<b$. First check for endpoint zeros of a function; otherwise require strictly opposite signs at the endpoints. Evaluate the midpoint and retain the half with strictly opposite endpoint signs, stopping immediately if the midpoint is a zero of a function. The intermediate value theorem ensures that every retained root bracket still contains a zero of a function. This is the construction in MIT's root-finding notes.^[ref]

For a worked square root calculation, use the polynomial $f(x)=x^2-2$ on $[1,2]$. Write $[a_k,b_k]$ for the root bracket after $k$ halvings, with $[a_0,b_0]=[1,2]$. The candidate evaluated during update $k$ is the old midpoint $m_{k-1}$:The initial midpoint is $m_0=1.5$. After four halvings the new midpoint is $m_4=1.40625$, and the bisection error bound is $|m_4-\sqrt{2}|\le 0.03125$. Its actual error is approximately $0.00796356237310$. The last tested candidate $m_3=1.4375$ and the reported midpoint $m_4$ are different points.

The useful feature of the bisection method is that the retained width halves on every update, independently of the shape of the function graph. The bisection error bound therefore supplies an accuracy guarantee before the zero of a function is known. No derivative is needed, and a valid initial root bracket gives convergence without a close initial guess. This makes the bisection method a dependable fallback for the faster Newton root-finding iteration.

For a requested absolute tolerance $\epsilon >0$, stop when the retained root bracket has width at most $2\epsilon$, and report its midpoint. The bisection error bound then guarantees $|m_k-x_{\ast }|\le \epsilon$ in exact arithmetic. Using the ceiling function and the base-two logarithm, the required number of halvings isThe zero allows stopping before any update when the initial root bracket is already sufficiently narrow. For the polynomial $f(x)=x^2-2$ on $[1,2]$ with $\epsilon =10^{-3}$, $k_{\min }=\lceil {\log }_{2}500\rceil =9$. After nine halvings,Thus the reported midpoint meets the tolerance; its actual error from the square root $\sqrt{2}$ is approximately $0.000825500126905$. After eight halvings the bound is $1/512>10^{-3}$, so eight do not give this worst-case guarantee.

Cache the endpoint values. Each update of the bisection method needs just one new function evaluation, at the old midpoint. With no early zero of a function, $k$ updates therefore cost $k+2$ function evaluations, including the two initial endpoints. Returning the new midpoint requires no additional function evaluation; requesting its value would add one. The nine-update example needs eleven evaluations for the guarantee.

A small function value $|f(m_k)|$ alone does not guarantee a small position error: rescaling $f$ changes those values without changing its zeros of a function. In a practical implementation, use an explicit width test and an iteration limit. With floating-point arithmetic, also stop if the computed midpoint equals either endpoint: the root bracket can no longer shrink, so repeating the update would loop forever. Report this precision limit separately if the requested tolerance remains unmet. Reliable sign evaluations are needed to preserve the root bracket; compare signs directly rather than multiplying endpoint values, which can suffer overflow or underflow in floating-point arithmetic.

For a rounded midpoint $m$ still inside a valid root bracket, the geometric error bound is $\max \{m-a_k,b_k-m\}$; it need not equal $W_k/2$ because of roundoff error. A rigorous computed enclosure also needs conservative arithmetic for that bound and reliable endpoint signs. The exact-arithmetic bisection error bound should not be claimed unchanged after precision prevents further halving.

For practical tolerance conventions, SciPy's bisect combines absolute and relative tolerances through $|x-x_0|\le \mathrm{xtol}+\mathrm{rtol}|x_0|$, and provides a maximum iteration count and convergence status. These differ from the purely absolute width test above.^[ref]

A bracketing method finds a zero of a function by maintaining a root bracket and shrinking it as new values of the function become available. The bisection method always tries the midpoint; its special attraction is a predictable bisection error bound without any derivative evaluations.

Composite Simpson's rule applies Simpson's rule to consecutive pairs of equal subintervals. Choose a positive even integer $n\ge 2$, put $h=(b-a)/n$, and take quadrature nodes $x_j=a+jh$ for $j=0,\dots ,n$. There are $n$ subintervals, $n/2$ two-subinterval panels, and $n+1$ quadrature nodes. ThenThe second sum is empty when $n=2$, recovering one panel. The quadrature weights, before multiplication by $h/3$, are $1,4,2,4,\dots ,2,4,1$. Every interior even-indexed quadrature node is an endpoint of two panels, so its two endpoint contributions add to two; every odd-indexed quadrature node is a midpoint with weight four. An odd $n$ cannot be covered by these pairs. The displayed formula requires equal spacing.^[ref]

For the exponential function on the interval $[0,1]$, the exact integral is $I=e-1$. With $n=2$, the formula gives $S_2\approx 1.718861151876593$, an overestimate of about $5.793234175\times 10^{-4}$. With $n=128$, it gives $S_{128}\approx 1.718281828494607$, an overestimate of about $3.556140064\times 10^{-11}$. The figure shows the intermediate refinements and compares every computed absolute error with the error bound for composite Simpson's rule.

For this exponential function, a geometric series also gives a compact independent formula. With $q=e^h$ and $h=1/n$, summing the $n/2$ panel contributions yieldsA Taylor series at $h=0$ givesThe Big O notation describes the remaining terms as $h$ tends to zero. Consequently $(S_n-I)/(S_{2n}-I)$ tends to sixteen: the computed ratios increase from about $15.65169$ to $15.99965$. Fourth-order truncation error is also obtained by Richardson extrapolation of the composite trapezoidal rule.^[ref]

Every allocation $(x,y)$ in the convex feasible region for a resource budget is a convex combination of its three extreme points:The condition $x+2y\le 6$ is exactly ${\lambda }_1\ge 0$; nonnegative $x,y$ give the other two nonnegative coefficients. Thus the half-space description and the convex hull description encode the same triangle. At $(2,1)$ all three coefficients are $1/3$, and at $(3,9/8)$ they are $1/8$, $1/2$, $3/8$. The green and orange points in the budget diagram therefore have two complementary interpretations: direct resource constraints and mixtures of extreme choices.

The three extreme points alone are not a convex set, because their midpoints are absent. Their convex hull fills the whole triangle. More generally, Caratheodory theorem bounds how many generating points are needed for one convex combination in a finite-dimensional real vector space; its geometric content is that a complicated collection can describe each attainable average with a short representation.

Consider a stipulated continuous-allocation problem. Amounts $x,y$ of two activities use $x+2y$ units of a common resource, with budget six. The feasible regionis a triangle with extreme points $(0,0)$, $(6,0)$ and $(0,3)$. It is the intersection of three closed half-spaces, hence a closed convex set. The general intersection argument is short: a line segment between two points in every constraint set belongs to every constraint set, so it belongs to their intersection.^[ref]

For an original numerical check, take $A=(1,2)$ and $B=(5,1/4)$. Their resource totals are $5$ and $11/2$. Every convex combination $Z(t)=(1-t)A+tB$ is nonnegative and has resource totalIn particular the midpoint is $(3,9/8)$, using $21/4$ units. This verifies the full line segment, rather than just a plotted sample of it.

This is useful whenever divisible quantities share linear limits: the feasible choices remain compatible under averaging. A new linear equation also preserves this property, because averaging two identical constraint values gives the same value. The affine subspace and closed half-space viewpoints explain this without requiring a separate argument for every budget.

A convex set permits every intermediate mixture of two of its points. If two choices meet its constraints, continuously averaging them also meets those constraints. This geometric property makes many feasible regions easier to explore: there is no forbidden gap along a straight move between feasible choices.

Formally, a subset $C$ of a real vector space is convex whenThe parameter $t$ traverses the entire line segment, including both endpoints. The definition does not require a closed subset, a bounded set, or a nonempty interior. The empty set, the whole vector space, and every affine subspace are convex; an open unit disk and a closed unit disk are both convex.^[ref]

The triangle inequality verifies the disk example directly: if $\Vert x\Vert ,\Vert y\Vert \le 1$, then $\Vert (1-t)x+ty\Vert \le (1-t)\Vert x\Vert +t\Vert y\Vert \le 1$. For the open disk, the last inequality is strict. The unit circle fails the same test: $(1,0)$ and $(-1,0)$ belong to it, but their midpoint $(0,0)$ does not. A curved route along the unit circle connects the endpoints, so being path-connected is weaker than being a convex set.

The distinction between a convex set and a convex function matters. The former concerns where one may choose a point; the latter concerns how a value changes along line segments. Convex optimization combines both: averaging preserves feasibility, and a convex function obeys an upper bound by the corresponding average of endpoint values. A convex set alone does not make an arbitrary objective a convex function or eliminate its inferior local minima.

Polynomial interpolation makes Simpson's rule exact for every polynomial of polynomial degree at most two. To understand the extra polynomial degree, centre the coordinate at the midpoint: $y=x-m$, with endpoints $-h,h$. For any cubic polynomial $p$, let $p_2$ be its interpolating quadratic polynomial. The polynomial $p-p_2$ vanishes at all three quadrature nodes, sowhere $A$ is the coefficient of $x^3$ in $p$. As a function of $y$, this is an odd function, whose integral over the symmetric interval $[-h,h]$ is zero. Hence integrating $p_2$ also integrates $p$ exactly. This is the symmetry mechanism in the nodal-polynomial criterion for quadrature exactness.

In the worked example, the polynomial difference factors asIts definite integrals on $[-1,1/2]$ and $[1/2,2]$ are respectively $81/64$ and $-81/64$. Their cancellation concerns signed integrals, rather than pointwise accuracy of polynomial interpolation.

The degree of precision of a quadrature rule stops at three: on the same interval, the monomial $x^4$ gives $S(x^4)=69/8$, whereas ${\int }_{-1}^2{x}^{4}dx=33/5$. Their difference, exact integral minus approximation, is $-81/40$.

De Casteljau's algorithm evaluates a Bézier curve by repeated linear interpolation between neighboring control points. Fix $t\in [0,1]$, start with $P_i^{(0)}(t)=P_i$, and computeEach new row has one fewer point. The final point is $P_0^{(n)}(t)=C(t)$. Use the same $t$ at every stage: it describes the fraction of each line segment traversed from its first endpoint toward its second. A cubic Bézier curve needs three levels containing $3$, $2$, and $1$ linear interpolations, respectively.^[ref]

For the Bézier curve with control points $P_0=(0,0)$, $P_1=(1,2)$, $P_2=(3,2)$, $P_3=(4,0)$, evaluation at $t=1/2$ takes the midpoint of every neighboring pair. The first level givesThe second level givesand the third level givesThe intermediate control polygons shrink toward the evaluated point. Each step is a convex combination, giving a geometric explanation of the convex hull property; every step also commutes with an affine map.

The same construction splits the Bézier curve exactly, with no approximation. The outer edges of the triangular array give the new control points: for a cubic, $(P_0,Q_0,R_0,S)$ specify the left piece and $(S,R_1,Q_2,P_3)$ specify the right piece.^[ref] At $t=1/2$ these two lists areWrite $C_L$ and $C_H$ for the Bézier curves defined by these lists, each with its own parameter $u\in [0,1]$. ThenTheir images reproduce the original curve, while their derivatives account for the shorter parameter intervals. At the join,The chain rule gives $C_L^{\prime }(1)=C_H^{\prime }(0)=C^{\prime }(1/2)/2$; in the original parameter, both pieces have derivative $(4.5,0)$. This is the interval scaling in the parametric first-derivative join of Bézier curves.

Repeated midpoint splitting is useful for drawing vector graphics paths: each piece has a smaller control polygon and its own convex hull enclosure. A renderer can refine pieces until they are sufficiently close to line segments at the chosen display scale. Evaluating $C(t)$ and splitting it are thus two results of the same short algorithm.

For a proof outline, expanding the recursion givesPascal's identity proves this by induction; at $r=n$, $i=0$ it is precisely the Bernstein basis formula for the Bézier curve. The geometric construction remains useful even when expanding the polynomial by hand would be cumbersome.

// Target: analysis.bigb

The parameter $e\ge 0$ in a focus-centred polar coordinates equation $r=\ell /(1-e\cos \theta )$ for a conic section. It distinguishes an ellipse ($e<1$), a parabola ($e=1$), and a hyperbola ($e>1$).

For a noncircular conic section, the eccentricity of a conic is the constant ratiowhere $F$ is a focus of a conic, $L$ the corresponding directrix of a conic, and $X$ a point on the curve. For an ellipse with semi-major axis $a$ and distance $c$ from the focal midpoint to either focus of a conic, $e=c/a=\sqrt{1-b^2/a^2}$. Its circle case has $e=0$; a finite directrix does not give that circle through the displayed ratio. The circle is the limit as $c\to 0$ and its directrices of a conic move arbitrarily far away.

The eccentricity of a conic is intrinsic to the cutting plane, and is not generally its slope in a chosen three-dimensional drawing. For the family $z=1+mx$ in conic section, with $0<m<1$, completing the square givesA displacement $dx$ along this plane has Euclidean norm $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, givingFor $m=3/5$, this is $3/\sqrt{17}$, rather than $3/5$. This normalization separates the curve's geometry from a distorted projection of it onto the $xy$ plane.

An extreme point of a convex set is a point that cannot be made by genuinely mixing two different points of the set. For a convex set $C$ in a real vector space, the condition on $p\in C$ isThus $p$ lies on no open line segment between two distinct points of $C$. It is enough to exclude midpoint representations: if $p$ lies inside a line segment, two nearby points on opposite sides have midpoint $p$.^[ref]

The square $[-1,1{]}^2$ has just four extreme points, at $(\pm 1,\pm 1)$. In a convex combination giving a corner, each coordinate is already at its largest or smallest allowed value, forcing both input points to agree with that corner. In contrast, the boundary point $(1,0)$ is not extreme:It is the midpoint of two other points on the same edge. An interior point also admits a small line segment through it.

The unit disk $D=\{z\in {\mathbb{R}}^2:\Vert z{\Vert }_2\le 1\}$ has a different shape: every point on its unit circle is extreme. For a unit vector $p$, the Cauchy-Schwarz inequality givesand equality is possible only at $z=p$. If a convex combination equals $p$, applying this linear functional forces both ingredients to attain that equality. Both must therefore be $p$. Extremality need not look like a sharp corner.

These points help describe all allowable mixtures. Every $(x,y)\in [-1,1{]}^2$ is a convex combination of its four corners, with the weight on $(a,b)\in \{-1,1{\}}^2$ equal to $(1+ax)(1+by)/4$. For the unit disk, write $z=rp$, with $0\le r\le 1$ and $p$ a unit vector; at zero choose any $p$. ThenTwo extreme points suffice for that mixture. The Krein-Milman theorem gives a much broader reconstruction: a nonempty compact convex set in a locally convex space whose topology is Hausdorff is the closed convex hull of its extreme points.

Mark Krein and David Milman published their reconstruction theorem in 1940. Milman's colleagues describe that result and its subsequent use as a method of representing convex sets through their extreme points.^[ref]

If the amounts in the budget example must be integers, the actual feasible region is $C\cap {\mathbb{Z}}^2$. Both $(0,0)$ and $(1,0)$ are feasible, but their midpoint $(1/2,0)$ is not. Thus the continuous feasible region is a convex set while the integer-restricted choices are not. Averaging two manufacturing plans may specify half an indivisible item. Integer programming must respect that restriction; using the surrounding convex set is a relaxation, not a proof that its averages are admissible integer plans.

Interval bisection solves a target equation $f(x)=t$ by repeatedly halving a closed interval known to contain a solution. It is the bisection method applied to $g(x)=f(x)-t$, so the target becomes a zero of a function. This reduction from an equation to root finding is described in Demanet's notes.^[ref] Here the emphasis is on a continuous strictly increasing function: the target identifies one input, making interval bisection a way to evaluate its inverse function without an explicit inverse formula.

Let $f:[a_0,b_0]\to \mathbb{R}$ be a continuous function, with $a_0<b_0$ and $f(a_0)\le t\le f(b_0)$. The intermediate value theorem guarantees existence; a strictly increasing function also guarantees uniqueness. Check endpoint equality first. Otherwise, for the current root bracket $[a_k,b_k]$, evaluate its midpoint $m_k=(a_k+b_k)/2$. If $f(m_k)<t$, retain $[m_k,b_k]$; if $f(m_k)>t$, retain $[a_k,m_k]$; if $f(m_k)=t$, return that exact solution. These comparisons preserve the target bracket. General bisection method needs continuity and opposite signs of $g$ at the endpoints, without requiring monotonicity or a unique zero of a function; see limitations of bisection.

An original dimensionless polynomial example isHere $f(0)=0<10<30=f(3)$ and the derivative $f^{\prime }(x)=3x^2+1>0$, so there is exactly one solution. In fact $x_{\ast }=2$; knowing this lets us check the computed absolute error independently of the guarantee. The first four halvings give the following exact rational numbers:

The beauty of interval bisection is that the enclosure itself supplies the certificate: one comparison removes half the possible inputs, regardless of the steepness of the function graph. In exact arithmetic, after $k$ completed halvings without exact termination, the width and the bisection error bound for the new midpoint areThe reported midpoint after update four is $m_4=63/32$, rather than the last tested point $m_3=33/16$. Its certified absolute error is at most $3/32$, and its actual absolute error is $1/32$. For a requested position tolerance $10^{-3}$, the bisection stopping criterion requires eleven halvings: ten give $3/2048>10^{-3}$, while eleven giveThe actual absolute error is $1/4096$. The new midpoint need not be evaluated to obtain this certificate.

The signed output residual is $r=f(m)-t$; the position approximation error is $m-x_{\ast }$. They answer different questions: how close is the output to the target, and how close is the input to the solution? Driscoll and Braun explain why a small residual needs information about the derivative to imply small position error.^[ref] For this particular polynomial, direct factorization givesThus at $m_4=63/32$ the residual is $-13121/32768\approx -0.400421$, despite position absolute error $1/32=0.03125$. If a differentiable function satisfies $0<\ell \le f^{\prime }(x)\le L$ throughout the bracket, the mean value theorem gives the rigorous conversionOur example permits $\ell =1$ and $L=28$ on $[0,3]$. Strict increase alone supplies no positive lower derivative bound: $f(x)=x^3$ is strictly increasing but $f^{\prime }(0)=0$. A small residual alone is therefore not a universal position certificate; even the target equation $10^{-6}(x-2)=0$ has residual $10^{-6}$ at $x=3$, a full unit from its solution.

The guarantees require continuity and correct evaluations, not just narrowing endpoints. For the strictly increasing function $h(x)=x$ when $x<0$ and $h(x)=x+1$ when $x\ge 0$, the target $1/2$ lies between $h(-1)$ and $h(1)$ but is never attained: a jump discontinuity defeats the existence argument. With floating-point arithmetic or an approximate evaluation, an incorrect comparison can discard the true solution. A certified interval enclosure of $f(m)$ wholly below or above $t$ certifies the corresponding decision; an enclosure containing $t$ is inconclusive, rather than proof of equality. If rounding makes the midpoint equal an endpoint, report the precision limit; see bisection stopping criterion for the rounded position bound. Both figures are exact mathematical examples, not measured data.

Several zeros of a function can lie in one initial root bracket: the bisection method seeks one, rather than enumerating them. For instance, the polynomial $(x-1)(x-2)(x-3)$ has opposite endpoint signs on $[0,4]$, but its first midpoint evaluation returns only the polynomial root $2$. The initial zero of a function need not be unique for the bisection error bound to apply to a retained zero of a function.

A zero of a function without a sign change cannot be located by the strict sign-change test alone. The polynomial $(x-1{)}^2$ touches zero at $1$ but is positive on either side. On $[0,3]$ both endpoint values are positive: although the closed interval contains a polynomial root, it fails the usual starting test for the bisection method. An endpoint or evaluated midpoint zero of a function can still be recognized directly.

Opposite signs certify a root bracket only with the required continuity. For $f(x)=1/x$, the values at $-1$ and $1$ have opposite signs, but $f$ is undefined at the midpoint $0$ and has no zero of a function. Hence $[-1,1]$ is invalid input for the bisection method; the intermediate value theorem does not apply across the gap.

The medial triangle joins the three midpoints of a triangle's sides. Each side is parallel to the corresponding original side and has half its length: $((\mathbf{x}+\mathbf{z})/2)-((\mathbf{x}+\mathbf{y})/2)=(\mathbf{z}-\mathbf{y})/2$. Each altitude of a triangle in the medial triangle is a perpendicular bisector of an original side, so its orthocenter is the original triangle's circumcenter.

Checking only midpoints is not enough for an arbitrary subset of a real vector space. The rational numbers contain the midpoint of every pair of rational numbers, but the line segment from $0$ to $1$ also contains irrational numbers. For instance $\sqrt{2}/2$ lies on that line segment and is not rational. This shows why the definition quantifies over every real $t\in [0,1]$.

There is a useful additional hypothesis: a closed subset $C\subseteq {\mathbb{R}}^d$ containing the midpoint of every pair of its points is a convex set. Repeated bisection includes all points $(1-t)x+ty$ with dyadic rational $t\in [0,1]$. These coefficients approach every real $t\in [0,1]$; being a closed subset then includes each limit. This is a short proof outline, and the rational numbers example identifies precisely why a closed subset was required.

For a short proof outline, endpoint values give the non-strict cases immediately. If $y$ is strictly between $f(a)$ and $f(b)$, apply the bisection method to the continuous function $g(x)=f(x)-y$. A midpoint with $g=0$ finishes the proof. Otherwise retain nested intervals $[a_k,b_k]$ with strictly opposite signs and widths $(b-a)2^{-k}$. The nested interval theorem, using completeness of the real numbers, gives a common point $c$ to which both endpoint sequences converge. After reversing the sign of $g$ if necessary, $g(a_k)<0<g(b_k)$. Continuity makes both endpoint value sequences converge to $g(c)$, forcing $g(c)\le 0$ and $g(c)\ge 0$, hence $f(c)=y$. The original endpoint values differ from $y$, so $c\in (a,b)$. This proof explains why a jump can defeat the construction: shrinking a candidate closed interval cannot supply the missing continuity.