The intermediate value theorem states that if $f:[u,v]\to \mathbb{R}$ is continuous and $y$ lies between $f(u)$ and $f(v)$, inclusive, then some $c\in [u,v]$ satisfies $f(c)=y$. If $y$ lies strictly between the endpoint values, then $c$ lies in $(u,v)$.

If there were no interior $x_2$ with $f^s(x_2)\le 0$, the symmetric derivative would be positive everywhere in $(a,b)$. Part ii would make $f$ strictly increasing there. Choose $a<u<v<b$. Continuity at the endpoints and monotonicity on the interior implycontradicting $f(a)=f(b)$. Hence such an $x_2$ exists.

If there were no interior $x_1$ with $f^s(x_1)\ge 0$, then $(-f{)}^s=-f^s>0$ everywhere. Apply the same argument to the continuous function $-f$, whose endpoint values are equal. This is again impossible. ThusThe two points need not be distinct.

We prove the symmetric derivative criterion for monotonicity using an antiderivative. Fix $a<u<v<b$ and defineThe fundamental theorem of calculus gives $F^{\prime }=f$ because $f$ is continuous. For each interior $x$,Since $f^s(x)>0$, the definition of symmetric derivative gives $\delta >0$ such that $f(x+t)-f(x-t)>t{f}^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,

We next show that $F$ is a convex function. On any closed interval $[r,s]\subset (a,b)$ let $\ell$ be the affine function joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the extreme value theorem would give a positive maximum at an interior point $x$. For every sufficiently small $h>0$, that maximum would implyThe affine function contributes zero to this symmetric difference, contradicting the strict positive inequality above. Thus $F\le \ell$ on every such interval, exactly convexity.

For a differentiable convex function, secant slopes bound endpoint derivatives. Explicitly, the chord inequality at $u+h$ and at $v-h$, followed by $h\downarrow 0$, givesSo $f$ is nondecreasing. If $f(u)=f(v)$ for some $u<v$, monotonicity would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is strictly increasing. This proof requires no continuity of $f^s$ and never applies the ordinary mean value theorem to a merely symmetrically differentiable function.

Write the symmetric difference quotient as the average of two ordinary difference quotients:If $f$ is differentiable at $x$, both terms tend to the derivative $f^{\prime }(x)$. Hence the symmetric derivative satisfies $\boxed{f^s(x)=f^{\prime }(x)}$.

The absolute value function $f(x)=|x|$ is continuous, and $f^s(0)=0$ because $|h|=|-h|$. Its right one-sided derivative at zero is $1$ and its left one-sided derivative is $-1$, so it is not differentiable at zero. Away from zero it is differentiable, hence has a symmetric derivative there as well. This supplies an example meeting the hypotheses on, for example, $(-1,1)$.

For any $u<v$ in the interval, the mean value theorem, applicable by differentiability implies continuity, givesAll pairs of values are equal. Thus $\boxed{f\text{ is a constant function}}$, the constancy theorem.

Take any $a<u<v<b$. By differentiability implies continuity, $f$ is continuous on $[u,v]$, and it is differentiable on $(u,v)$. The mean value theorem gives $c\in (u,v)$ for whichTherefore $\boxed{f(u)\le f(v)\text{ whenever }u<v}$: $f$ is a nondecreasing function, which is the non-strict meaning of increasing here.

The mean value theorem states: if a real-valued function $f$ is continuous on a closed interval $[u,v]$, where $u<v$, and differentiable on $(u,v)$, then there is $c\in (u,v)$ with

A function $f:(a,b)\to \mathbb{R}$ is differentiable on $(a,b)$ if, for every $x\in (a,b)$, the finite limit of a functionexists. Here $h\ne 0$ and is sufficiently small that $x+h\in (a,b)$. This limit of a function is the derivative of $f$ at $x$.

Start with all of $N_0$ and $N_1$, namely $-1,-1/2,-1/2$, whose sum is $-2$. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the tripleThis uses all $2^{n+1}$ occurrences of $P_n$ and all $2^{n+2}$ occurrences of $N_{n+2}$, exactly once. Together with the initial terms, every original occurrence is used, establishing a rearrangement of a series. Its beginning isEach triple has sum zero. At its intermediate terms the partial sum is at most $2^{-(n+1)}$ above $-2$. This bound tends to zero, so the whole series, rather than just the subsequence of sums at completed triples, has sum $\boxed{-2}$.

Place the single term in $N_0$ first. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the pairThis uses all of $P_n$ and all of $N_{n+1}$ exactly once, so it is an explicit rearrangement of a series. Its beginning isEvery completed pair leaves the partial sum at $-1$. Within a level-$n$ pair the partial sum differs from $-1$ by at most $2^{-(n+1)}$. As $n\to \infty$ this tends to zero, proving that the rearranged series has sum $\boxed{-1}$.

In level $n$, each group isand there are $2^n$ such groups. The partial sums at completed groups are therefore zero. Inside a group of level $n$, the only other partial sums are $2^{-(n+1)}$ and $2^{-n}$. Every level contains finitely many terms, so the level containing the $N$th term tends to infinity with $N$. Thus all partial sums lie between zero and a bound tending to zero. By the squeeze theorem, the series has sum $\boxed{0}$.

For the next two parts, label the positive occurrences of size $2^{-(n+1)}$ by $P_n$ and the negative occurrences of size $-2^{-n}$ by $N_n$. There are exactly $2^{n+1}$ occurrences in $P_n$ and $2^n$ in $N_n$. Occurrences with equal value retain separate labels, so checking their counts verifies a bijection of terms.

Let $\epsilon >0$. By the definition of Cauchy sequence, choose an integer $K\ge 1$ such that $|x_j-x_k|<\epsilon /2$ whenever $j,k\ge K$. Apply the given inequality with $N=K+1$. It supplies $n>K+1$ for whichBoth $n$ and $K$ are in the Cauchy sequence estimate, soFor every $m\ge K$ the triangle inequality givesThis is the definition of convergence to zero: $\boxed{x_m\to 0}$. The proof uses only the Cauchy sequence estimate, positivity, and the stipulated inequality; it does not assume the general completeness of the real numbers theorem for Cauchy sequences. Taking $N=K+1\ge 2$ also avoids the empty initial set in a literal reading of the condition for very small $N$.

If $L<1$, set $\epsilon =(1-L)/2>0$. The definition of limit of a sequence gives $a_n<L+\epsilon =(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\ge 1$. Hence $\boxed{L\ge 1}$.

Factor the difference of cubes. Since both cube roots are at least one,Given $\epsilon >0$, the definition of limit of a sequence supplies $N$ with $|a_n-L|<3\epsilon$ for $n\ge N$. The displayed inequality then gives $|a_n^{1/3}-L^{1/3}|<\epsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.

False. Take the finite sets $A=B=\{-2,-1\}$. Their suprema are both $-1$, whereas the set of pairwise products is $A\cdot B=\{1,2,4\}$. ConsequentlyNegative factors can reverse an inequality, which prevents the proposed argument with separate suprema.

True. Write $\alpha =\sup A$ and $\beta =\inf B$. Every $a\in A$, $b\in B$ satisfies $a-b\le \alpha -\beta$, giving an upper bound for the set $A-B$ of pairwise differences. For every $\epsilon >0$, the definitions of supremum and infimum give $a\in A$ and $b\in B$ withThen $a-b>\alpha -\beta -\epsilon$. Hence no smaller upper bound works, andHere $A-B$ means pairwise subtraction, rather than set difference.

True. Put $\alpha =\sup A$, $\beta =\sup B$ and $u=\max \{\alpha ,\beta \}$. These suprema exist by completeness of the real numbers, since the sets are nonempty and bounded. Every set element of the union $A\cup B$ is at most $u$, so $u$ is an upper bound. Conversely, every upper bound of $A\cup B$ bounds both $A$ and $B$, and is therefore at least both suprema. Thus

For a nonabelian group, iv and iii imply $|G|=|G/A||A|\ge 4|A|$. Substituting $|A|\le |G|/4$ into i yieldsThe constant is sharp: the quaternion group has eight elements and the five conjugacy classes $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$, $\{k,-k\}$. To verify this directly, $\pm 1$ are central elements, conjugation by $j$ sends $i$ to $-i$, and similarly for $j,k$; its relations $i^2=j^2=k^2=-1$ and $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$ show that conjugating $i,j,k$ by any of $\pm 1,\pm i,\pm j,\pm k$ changes only its sign. Hence $m(G)=5=(5/8)|G|$.

A conjugacy class $C(x)$ is a singleton exactly when $gx{g}^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the singleton conjugacy classes correspond exactly to elements of the center of a group $A$, and there are $|A|$ of them.

Every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, if $r$ is the number of these remaining classes then $2r\le |G|-|A|$. ConsequentlyThis conjugacy class count bound from the center needs no classification of the finite group.

If $G$ is not Abelian, the contrapositive of ii says that $G/A$ is not a cyclic group. The quotient group is finite because $G$ is finite. By iii, every group with fewer than four elements is a cyclic group. HenceEquivalently, the center of a group satisfies $|A|\le |G|/4$, by the coset count in iii.

A smallest example is the Klein four-groupwith coordinatewise addition modulo $2$. Every nonidentity element has order of a group element $2$, so none generates all four elements; hence this group is not a cyclic group.

A group of group order $1$ is a cyclic group. A group of group order $2$ or $3$ is also a cyclic group: choose a nonidentity element; by Lagrange's theorem, its order of a group element is a nontrivial divisor of that prime number, so equals the whole group order. Thus no smaller example exists, and the smallest possible group order is $4$.