Sturm signed remainders count distinct real polynomial roots

Each nonzero remainder has degree smaller than its divisor by [F1], so the chain terminates. The Euclidean algorithm preserves the common divisors of $f$ and $f^{\mathrm{\prime }}$, and their gcd is $1$ by the square-free hypothesis; hence the last nonzero member is a nonzero constant. A nonzero polynomial has only finitely many real roots: repeatedly divide by $t-c$ at each distinct root $c$ in $\mathbb{R}[t]$; each division lowers degree, so there can be no more roots than the original degree.

For each intermediate index $0<j<m$, the division identity is $f_{j-1}=q_j{f}_j-f_{j+1}$. If $f_j(c)=0$, then $f_{j-1}(c)=-f_{j+1}(c)$. No two consecutive members can vanish at one point: the identity would propagate a common zero to the final nonzero constant. Thus both neighbors are nonzero and have opposite signs at $c$.

A polynomial $h$ with $h(c)\mathrm{\ne }0$ keeps its sign in some interval about $c$: division by $t-c$ gives $h(t)=h(c)+(t-c)k(t)$, and the finite polynomial $k$ is bounded on $[c-1,c+1]$ by the sum of the absolute values of its coefficients times the corresponding powers of $\mathrm{\mid }c\mathrm{\mid }+1$. Taking $\mathrm{\mid }t-c\mathrm{\mid }$ small enough makes $\mathrm{\mid }(t-c)k(t)\mathrm{\mid }<\mathrm{\mid }h(c)\mathrm{\mid }$. Therefore, near a zero $c$ of an intermediate $f_j$, its two neighboring members keep their opposite signs; whether $f_j$ is positive, negative, or zero, those three entries contribute exactly one sign change. Intermediate zeros do not change $V$, including at an endpoint where $f_0$ is nonzero.

If $c$ is a root of $f_0=f$, then $f_1(c)=f^{\mathrm{\prime }}(c)\mathrm{\ne }0$: otherwise the Euclidean algorithm and [F1] give polynomials $u,v\in K[t]$ with $uf+v{f}^{\mathrm{\prime }}=1$, and evaluating at $c$ would give $0=1$. Division in $\mathbb{R}[t]$ gives $f(t)=(t-c)h(t)$ by [F6]. Expanding $h$ in powers of $t-c$, the coefficient of $(t-c)$ in $f$ is $h(c)$; the formal derivative formula [F2] says this coefficient is $f^{\mathrm{\prime }}(c)$. By step 3.1, $h$ keeps the sign of $f^{\mathrm{\prime }}(c)$ near $c$, so $f$ changes sign exactly once there, from the opposite sign to the same sign as $f^{\mathrm{\prime }}(c)$. The first pair $(f_0,f_1)$ therefore loses exactly one sign change as $t$ passes left to right through $c$.

The chain has finitely many roots by step 1.1. Every chain member is a real polynomial and so is continuous by [F4]; if its sign changed between two consecutive zeros, [F5] would give an additional zero. Thus between the finitely many chain-root locations all signs are fixed by [F6]. At every intermediate-member zero the variation is unchanged by step 3.1, and at every root of $f$ it decreases by one by step 4.1. Since $f(a)f(b)\mathrm{\ne }0$, neither endpoint is a root of $f$; summing the changes from $a$ to $b$ gives $V(a)-V(b)$ equal to the number of distinct roots of $f$ in $(a,b)$.