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True. Rationalizing the square root gives, for $n\ge 1$,The uniform bound is exactly the assertion $O(1)$ in Big O notation.

Use the intended first-order structure of real arithmetic, with Int selecting the integers; $z$ is the parameter in ${\varphi }_z$. A suitable first-order formula isThe first condition requires a real square root of $2z$. The second excludes every representation as a rational number $p/q$ with a nonzero integer denominator. This uses only the listed function and constant symbols, the Int predicate and logical equality; in particular, $2$ is written $1+1$ and there is no extra division symbol. A square root exists exactly when $z\ge 0$, and either sign of a root is a rational number exactly when the other sign is a rational number, so the absence of a positivity condition on $r$ does not affect the claim about the principal square root. At $z=0$ the formula is false; for $z<0$ there is no real witness. The meanings of $+$ and $\ast$ as real arithmetic, not just the meaning of Int, are part of this interpretation.

A Pauling bond-energy excess subtracts a chosen like-atom baseline from the unlike-atom bond-dissociation energy. For an unequal-input arithmetic example, stipulate $D_{AA}=200$, $D_{BB}=450$ and $D_{AB}=421.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, all with one dissociation convention. The arithmetic mean baseline and excess areThe arithmetic-baseline Pauling scale model therefore gives $|{\chi }_{\mathrm{P}}(A)-{\chi }_{\mathrm{P}}(B)|=\sqrt{96.485/96.485}=1$. The bond-dissociation energy $421.485$ is not itself the excess: subtracting the baseline is essential. All three numbers are stipulated, not experimental values for named chemical elements.

Changing the baseline changes the model. For the same stipulated inputs, a geometric mean baseline would be $\sqrt{200\times 450}=300\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, yielding an excess of $121.485\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and a difference of approximately $1.122$. It is a different construction, even though both start from the same bond-dissociation energies. State which baseline is used before comparing electronegativity differences.

The arithmetic-baseline Pauling scale also has a useful inverse: a difference of $0.5$ corresponds to an excess of $24.12125\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a difference of $2$ corresponds to $385.94\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Doubling the electronegativity difference quadruples the modeled excess. Interchanging the endpoint labels changes neither quantity; the square root alone cannot choose the direction of bond polarization.

In The Nature of the Chemical Bond. IV. The Energy of Single Bonds and the Relative Electronegativity of Atoms (printed pp. 3576–3578, Table II and equation (1)), Linus Pauling organized excess energies for chemical bonds into an approximately additive map of their square roots. He also identified a limitation: his H–F map separation of $2$ predicted an excess of $4.00\mathrm{eV}$, exceeding his table's value by $1.23\mathrm{eV}$. Those are historical values and a historical model discrepancy, not modern recommended inputs.^[ref]

For a nonnegative integer $n$, compute its integer square root $r=\lfloor \sqrt{n}\rfloor$ and accept exactly when $r^2=n$. An approximate square root must be corrected with exact integer comparisons before this equality is tested. This procedure distinguishes a squareness test from merely approximating a square root.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.

Put $I=I(1)$. The nonnegative integrand allows Tonelli's theorem to write the square as a double integral:In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant supplies the area factor $r$:Since $I>0$, take the positive square root to get $I=\sqrt{\pi }$. The computation proves finiteness as well as the value. For an explicit limiting argument, first integrate over disks of radius $R$; their integrals are $\pi (1-e^{-R^2})$, which increase to $\pi$ as $R\to \infty$. ^[ref]

The geometric surprise is that radial shells get longer as $r$ increases, even while the Gaussian function gets smaller. Its central height is largest at zero, but the radial contribution $2\pi r{e}^{-r^2}$ vanishes there and peaks at $r=1/\sqrt{2}$. Differentiating gives $2\pi e^{-r^2}(1-2r^2)$, which changes sign at that radius. Keeping the Jacobian determinant is what makes the one-dimensional integral and the plane geometry agree.

A prime number is a positive integer $p>1$ whose only positive divisors are $1$ and $p$. The first ten prime numbers are $2,3,5,7,11,13,17,19,23,29$. In contrast, $12=3\cdot 4$ is a composite number because it has a positive divisor strictly between one and itself. The integer one is neither a prime number nor a composite number: the lower bound $p>1$ is part of the definition. Equivalently, the divisor function satisfies $\tau (p)=2$ for a prime number, whereas $\tau (1)=1$ and $\tau (n)\ge 3$ for every composite number $n$.

Prime numbers are the building blocks of positive integers under multiplication. The fundamental theorem of arithmetic says that every integer greater than one is a product of prime numbers, with a unique collection of factors when their order is ignored. For example, $12=2\cdot 2\cdot 3$ and $18=2\cdot 3\cdot 3$. Their common factors reveal the greatest common divisor $6$, and the combined maximum powers reveal the least common multiple $36$. A complicated divisibility question becomes a comparison of prime exponents.

There is also a useful finite test. If a composite number $n$ equals $ab$ with $1<a\le b$, then $a^2\le n$. Therefore some prime factor is at most $\sqrt{n}$, and trial division only needs to check prime numbers up to that square root. For $97$, checking $2,3,5,7$ suffices; none divides $97$, so it is a prime number. For $91$, the same test finds $7$, giving $91=7\cdot 13$. To find many prime numbers together, the sieve of Eratosthenes reuses the work by marking multiples.

The division with remainder calculations for $97$ make the test explicit:Since $9^2<97<10^2$, these are all the prime numbers at most its square root, and every remainder is nonzero. Thus the calculation certifies a prime number, without testing the integers $8,9,10,\dots ,96$ individually. In contrast, $91=7\cdot 13+0$ gives a prime factor and proves that $91$ is a composite number.

The definition has a long history. Book VII of Euclid's Elements distinguishes prime numbers from composite numbers through their possible measures; Book IX, Proposition 20 establishes that no finite list exhausts the prime numbers. The modern argument under Euclid's theorem needs only divisibility, so it proves an infinite conclusion without estimating how frequently prime numbers occur. ^{[ref] [ref]}.

For an odd prime $p$, a nonzero congruence class $a$ is a quadratic residue when $a\equiv x^2(\mathrm{mod}p)$ for some integer $x$. Here "quadratic residue" excludes zero: zero is a square of a number, but is neither a quadratic residue nor a quadratic nonresidue under this convention. Some treatments include zero among quadratic residues; the distinction matters when counting them or discussing their subgroup.^[ref]

Modulo seven, squaring the six nonzero congruence classes givesThus the set of quadratic residues is $Q_7=\{1,2,4\}$, and the quadratic nonresidues are $\{3,5,6\}$. In particular, $x^2\equiv 2(\mathrm{mod}7)$ has exactly the two solutions $x\equiv 3,4(\mathrm{mod}7)$: $3^2=9\equiv 2$ and $4^2=16\equiv 2$. Modulo eleven the quadratic residues are instead $Q_{11}=\{1,3,4,5,9\}$. Two is therefore a quadratic residue modulo seven and a quadratic nonresidue modulo eleven: being a quadratic residue depends on the modulus.

Why do the square roots come in pairs? If $x^2\equiv y^2(\mathrm{mod}p)$, the prime number $p$ divides $(x-y)(x+y)$, so it divides one factor: $y\equiv x$ or $y\equiv -x(\mathrm{mod}p)$. For nonzero $x$ these are distinct, because $p$ is an odd prime. Every nonempty nonzero fiber of a function of $s(x)=x^2$ therefore contains exactly two inputs. Pairing the $p-1$ nonzero congruence classes gives exactly $(p-1)/2$ quadratic residues and $(p-1)/2$ quadratic nonresidues. Including zero gives $(p+1)/2$ different squares, with one rather than two square roots for zero.

The quadratic residues form a subgroup of the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$, with index of a subgroup two. Indeed, one is a square, $(u^2)(v^2)=(uv{)}^2$, and $(u^2{)}^{-1}=(u^{-1}{)}^2$; the inverses exist because the inputs are nonzero in the prime field. Equivalently, the squaring group homomorphism has kernel of a group homomorphism $\{1,-1\}$ and image of a function $Q_p$. The apparent loss of a sign in squaring becomes a precise division of the nonzero arithmetic into two equal parts.

This structure turns a search for square roots into a yes-or-no test: Euler's criterion detects quadratic residues through modular exponentiation, and the Legendre symbol records the result as a sign. For a general integer modulus, the modular congruence $x^2\equiv a(\mathrm{mod}n)$ still makes sense, but the two-root count and the prime field argument need not survive. For example, $x^2\equiv 1(\mathrm{mod}15)$ has four solutions $1,4,11,14$; the Chinese remainder theorem separates the sign choices modulo three and five.

For $u,v\ge 0$ with $u+v>0$, multiplication by the sum of the square roots givesThis identity exposes small differences hidden by cancellation. For example, $\sqrt{n^2+n}-n=1/(\sqrt{1+1/n}+1)\to 1/2$. It also converts normalized square-root increments into terms comparable with the harmonic series.

Multiplying numerator and denominator of a fraction by the same nonzero expression can remove a square root from the denominator. For instance,when $a\ge 0$ and both denominators are nonzero. This uses the difference-of-squares identity $(\sqrt{a}+b)(\sqrt{a}-b)=a-b^2$. If $a=b^2$, the original expression may still be defined even though the rationalized quotient is not, so that case must be handled separately. The related rationalization of a square-root difference applies the same multiplication to a numerator.

A nonnegative integer $M$ is a triangular number exactly when $8M+1$ is a perfect square. Its square root is then an odd integer, and the row index isCompleting the square in $M=n(n+1)/2$ proves necessity; reversing the calculation proves sufficiency. For example, $8\cdot 21+1=169=13^2$, giving $n=6$. In contrast, $8\cdot 20+1=161$ lies strictly between $12^2$ and $13^2$, so twenty is not a triangular number. The criterion turns an apparently open-ended search through a sequence into one exact perfect square check.

The root mean square of a real function or data set is the square root of its mean square. A sinusoid of amplitude $a$ has root mean square $|a|/\sqrt{2}$. For orthogonal or uncorrelated zero-mean components, the squared root mean square of their sum is the sum of their mean squares; this follows by expanding the square and observing that all cross terms average to zero.

Setting $a(x-h{)}^2+k=0$ gives $(x-h{)}^2=-k/a$. A positive $-k/a$ gives two distinct roots of a polynomial in the real numbers; zero gives a repeated root of a polynomial at $h$; a negative value gives a pair of complex conjugate roots of a polynomial. For the downward-opening example, the two roots of a polynomial are $2\pm \sqrt{3/2}$, approximately $0.775255$ and $3.224745$.

The discriminant $\Delta =b^2-4ac$ packages the same calculation, becauseSince $4a^2>0$, the sign of $\Delta$ determines which of the three cases occurs. This derives the quadratic formula by a polynomial identity, and explains its structure: first shift to $h$, then take a square root. For the upward-opening example, $(x+3{)}^2=-4$ gives the complex conjugate roots of a polynomial $-3\pm 2i$.

The singular values of a matrix measure stretching of Euclidean norm. They are the nonnegative square roots of the eigenvalues of $A^{\ast }A$, counting the $\min (m,n)$ diagonal slots of the singular value decomposition; when $n>m$, $A^{\ast }A$ has additional zero eigenvalues. Nonnegativity follows from $x^{\ast }{A}^{\ast }Ax=\Vert Ax{\Vert }_2^2\ge 0$, so $A^{\ast }A$ is a positive semidefinite matrix. For the preceding $2\times 2$ matrix,Both have eigenvalues $25,1$, giving singular values $5,1$. In particular,Here the operator norm uses Euclidean norm: no unit vector stretches more than fivefold. Zero singular values identify lost directions; the number of positive singular values equals the matrix rank. This summarizes both amplification and information loss with nonnegative numbers.^[ref]

Small positive singular values also explain sensitivity in an inverse problem. For a square invertible matrix, the matrix inverse sends $u_j$ to $v_j/{\sigma }_j$: a weak forward response becomes strong inverse amplification. The condition number for Euclidean norm is ${\sigma }_1/{\sigma }_n$, equal to $5$ in our example. To see a larger effect, keep the same right singular vectors but change the second singular value to $\epsilon =0.01$:The output perturbation has Euclidean norm $0.001$, but the recovered input changes by $0.1$. Its hundredfold amplification follows from the direction of the perturbation, not just its size. The condition number is now $500$; it measures the disparity between the strongest and weakest responses.^[ref]

For repulsive contact interactions in a three-dimensional isotropic harmonic trap, neglecting the kinetic term in the Gross–Pitaevskii equation gives $n(r)=(\mu -m{\omega }^2{r}^2/2)/U_0$ inside $R=\sqrt{2\mu /(m{\omega }^2)}$ and zero outside. Normalization yields $N=8\pi \mu R^3/(15U_0)$, hence $\mu =(\hslash \omega /2)(15Na/a_0{)}^{2/5}$ and $R=a_0(15Na/a_0{)}^{1/5}$, with oscillator length $a_0=\sqrt{\hslash /(m\omega )}$. The density is parabolic but the wavefunction is its square root. Neglected kinetic energy rounds the edge and supplies a tail.

On $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choose the holomorphic branch $f(z)=\sqrt{1-z}\sqrt{1+z}$, taking each factor's principal square root. This gives $f(x)>0$ for $-1<x<1$. Across the branch cuts, its limits are $f(x+i0)=-i\sqrt{x^2-1}$ and $f(x-i0)=+i\sqrt{x^2-1}$ for $x>1$, and $f(x+i0)=+i\sqrt{x^2-1}$ and $f(x-i0)=-i\sqrt{x^2-1}$ for $x<-1$. The endpoints are branch points; analytic continuation around a single endpoint changes the square-root sign.

The map $z\mapsto z^2$ is a bijective conformal map from the open first quadrant to the complex upper half-plane. In polar coordinates it doubles arguments from $(0,\pi /2)$ to $(0,\pi )$, and its derivative $2z$ never vanishes inside the quadrant. Its inverse is the branch of the square root with argument in $(0,\pi /2)$. By conformal invariance of harmonicity, a harmonic function $f$ on the half-plane pulls back to $g(x,y)=f(x^2-y^2,2xy)$. The two boundary rays map to the positive and negative real axes, respectively.

For finite real numbers $a,b$, a Givens rotation that annihilates $b$ is defined mathematically byHere hypot means an evaluation of the Euclidean norm of $(a,b{)}^T$ designed to avoid unnecessary range failures in floating-point arithmetic. The zero case selects the identity matrix, avoiding division by zero. If $a=0,b\ne 0$, then $c=0,s=\mathrm{sign}(b)$; if $b=0,a\ne 0$, then $c=\mathrm{sign}(a),s=0$. Thus the retained component is nonnegative, including when $a<0$. In exact arithmetic, $ca+sb=r$ and $-sa+cb=0$.

Computing the square root of a*a + b*b directly can fail even when the final norm is representable. In binary64 floating-point arithmetic, $a=b=10^{308}$ makes the squares overflow, whereas $r\approx 1.4142\times 10^{308}$ is finite. With $a=b=10^{-300}$ the squares round to zero, although $r\approx 1.4142\times 10^{-300}$ is representable. A range-safe hypot avoids these intermediate failures.

The simple divisions $a/r,b/r$ still need care when $r$ is a subnormal number: rounding $r$ can cause large relative errors. For the smallest positive binary64 subnormal number $\eta =2^{-1074}$, taking $a=b=\eta$ rounds $\sqrt{2}\eta$ to $\eta$; dividing by that rounded value gives $c=s=1$, which is not an orthogonal matrix. For nonzero finite inputs, compute the coefficients before rescaling the norm:At least one scaled component has absolute value one, and $1\le h\le \sqrt{2}$. This construction keeps the coefficients meaningful even when rounding $r$ is severe. It cannot make an unrepresentably large $r$ finite, preserve an arbitrarily tiny $s$ against underflow, or repair inputs already rounded to zero. The row updates themselves can also overflow if their exact results exceed the available range.

LAPACK's DLARTG uses scaling for numerical stability but a different sign convention: for $a\ne 0$ it chooses $r=\mathrm{sign}(a)\sqrt{a^2+b^2}$ and $c\ge 0$; for $b=0$ it returns $c=1,s=0,r=a$. Its block still has the lower-left minus sign. Do not combine its returned coefficients with an assumed nonnegative $r$.^[ref]

A Hermitian matrix $A$ is a positive-definite matrix exactly when its leading principal minorsare all positive. The leading principal minor uses the first $k$ rows and the first $k$ columns in the same ordering. For the two-by-two Cholesky decomposition example, ${\Delta }_1=4$ and ${\Delta }_2=4\cdot 5-2^2=16$, so the criterion proves positive definiteness without calculating eigenvalues.

The successive pivot elements before taking square roots are ${\Delta }_k/{\Delta }_{k-1}$, with ${\Delta }_0=1$. Here they are $4$ and $16/4=4$. In general, a leading triangular matrix factor has determinant equal to the product of its diagonal entries, soConversely, positive leading principal minors make each successive Schur complement pivot element positive, allowing the Cholesky decomposition to be constructed. The identity $u^{T}Au=\Vert L^{T}u{\Vert }_2^2$ then proves positive definiteness; for complex matrices, replace the transposes by conjugate transposes. This gives a short inductive proof of the criterion.

For $B=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, the first minor is positive but $\det B=-3$. The second Cholesky radicand would be $1-2^2=-3$, and $v=(1,-1{)}^T$ directly gives $v^{T}Bv=-2<0$. Thus positive diagonal entries are insufficient. Replacing “positive” by “nonnegative” in just the leading-minor test is also insufficient for positive semidefiniteness: $\mathrm{diag}(0,-1)$ has leading minors $0,0$, but its quadratic form is negative on $(0,1{)}^T$.

A Taylor polynomial replaces a function near a chosen point by a polynomial with matching derivatives. If $f$ has derivatives through order $n$ at $a$, defineThe factorial denominator makes $T_{n,a}^{(j)}(a)=f^{(j)}(a)$ for $0\le j\le n$. This determines a unique polynomial of degree of a polynomial at most $n$. Order zero matches the value, order one also matches the tangent line, and order two also incorporates the second derivative. Matching more local information is useful because polynomials are easy to evaluate, differentiate, and integrate. The distinction between the finite Taylor polynomial and the infinite Taylor series is developed in OpenStax, Taylor and Maclaurin series.

For a computed example, let $f(x)=\sqrt{1+x}$ and expand at zero. Its first three derivatives giveAt $x=0.1$, the linear approximation is $1.05$ and the quadratic approximation is $1.04875$, compared with $\sqrt{1.1}\approx 1.048808848$. The quadratic correction captures the downward bending of the square root and reduces the absolute difference from about $0.00119115$ to $0.0000588482$.

The Taylor remainder turns this convenient approximation into a controlled calculation. On $[0,0.1]$, $f^{\prime \prime \prime }(x)=3/[8(1+x{)}^{5/2}]$ is positive and at most $3/8$. The Taylor theorem with Lagrange remainder therefore givesConsequently $1.04875<\sqrt{1.1}\le 1.0488125$ follows without using the displayed decimal evaluation as evidence. The Taylor polynomial and the Taylor remainder play different roles: the former supplies an answer that is easy to compute, while the latter specifies what accuracy is justified. Farther from the expansion point, the same low-degree polynomial need not remain accurate.