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Truncation toward zero removes the fractional magnitude of a real number, giving the integer obtained by applying the floor function to its absolute value and restoring the original sign; a magnitude below one gives zero. More precisely,The result is zero for $-1<x<1$. For example, $\mathrm{trunc}(2.7)=2$ and $\mathrm{trunc}(-2.7)=-2$, whereas $\lfloor -2.7\rfloor =-3$. Python's math.trunc uses this towards-zero convention, so replacing it by math.floor changes calculations with negative noninteger inputs. ^[ref].

For every real number $x$,The discarded difference $x-\mathrm{trunc}(x)$ has the sign of $x$ when $x$ is noninteger. In contrast, the fractional part $x-\lfloor x\rfloor$ is always in $[0,1)$. At $x=-2.7$, these two differences are $-0.7$ and $0.3$. Thus specifying a signed fractional remainder requires specifying which integer-valued operation is used.

Unlike the floor function and ceiling function, truncation toward zero need not commute with adding an integer. At $x=-0.7$ and $j=1$, $\mathrm{trunc}(x+j)=\mathrm{trunc}(0.3)=0$, whereas $\mathrm{trunc}(x)+j=1$. This matters if negative coordinates are assigned to equal-width bins by converting a quotient to an integer: $x=-0.7$ belongs to the bin $[-1,0)$ under the floor function, but truncation gives zero. The appropriate operation follows from the declared bin intervals.

Truncation toward zero is a nondecreasing function with a discontinuity of a function at each nonzero integer. It has no discontinuity of a function at zero, because it is identically zero on $(-1,1)$. Its agreement with the floor function on the positive side and the ceiling function on the negative side explains the different closed endpoints in the diagram.

A sequence of functions $f_n:X\to Y$ converges uniformly to $f:X\to Y$, where $Y$ is a metric space with metric $d$, precisely whenFor real numbers, use the absolute value $d(u,v)=|u-v|$. Compare the universal quantification and existential quantification:The decisive change is whether $N$ can depend on $x$. Uniform convergence implies pointwise convergence.^[ref]

For functions taking values in real numbers on a nonempty domain of a function, the supremum error is $E_n={\sup }_{x\in X}|f_n(x)-f(x)|$, allowing $+\infty$. Then uniform convergence is equivalent to $E_n\to 0$. The supremum need not be a maximum: pointwise strict bounds below $\epsilon$ only give $E_n\le \epsilon$, so use tolerance $\epsilon /2$ when proving this equivalence. This single error bound controls every possible input, including one chosen after seeing the index.

For $f_n(x)=x^n$ and its pointwise limit on $[0,1]$, the error is zero at one but approaches one as $x\to 1$ from below. Consequentlyfor every positive integer $n$. This supremum is never attained: below one the error is $x^n<1$, and at one it is zero. Deleting the endpoint leaves the same supremum error on $[0,1)$. To disprove uniform convergence directly, choose $\epsilon =1/2$: for any proposed $N$, take $n=N$ and $x=2^{-1/N}<1$. The error equals $1/2$, violating the strict tolerance. Thus a moving input defeats every proposed common threshold.

On the smaller interval $[0,a]$, for a fixed $0\le a<1$, the pointwise limit is zero andFor $0<a<1$ and $0<\epsilon <1$, the exact first successful common threshold isThe logarithm and floor function turn an infinite approximation statement into a finite stopping rule. At $a=0.8$, $0.8^{20}=0.01152921504606846976$ fails tolerance $0.01$, while $0.8^{21}=0.009223372036854775808$ succeeds. At $a=1/2$ and $\epsilon =1/8$, the ratio is exactly three: $N=4$ is necessary because equality is insufficient. For $a=0$, every error is zero. The domain of a function determines whether one threshold can certify every input.

For another explicit stopping rule, let $h_n(x)=x/(n+1)$ on $[0,1]$. Its uniform limit is zero and its supremum error is $1/(n+1)$. For every $\epsilon >0$, $N=\max \{1,\lceil 1/\epsilon \rceil \}$ works, using the ceiling function. Unlike $x^n$ on $[0,1]$, it controls even the endpoint.

The practical beauty of uniform convergence is that one quantitative bound can survive further operations: the uniform limit theorem preserves continuity, and uniform convergence preserves integrals on a compact interval controls accumulated error. Differentiation needs additional hypotheses, such as uniform derivative convergence with an anchored value. These results distinguish an approximation that looks accurate at selected inputs from one that justifies replacing a whole function in a calculation.