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The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

A sequence $(a_n)$ of real numbers is a convergent sequence with limit of a sequence $\ell$ whenDivergence to positive infinity meansIn both definitions the same $N$ must control every later term. The second condition is not convergence to a real number.

For a series of complex numbers, form the partial sums $S_N={\sum }_{n=1}^N{a}_n$. The series converges when these partial sums have a finite limit of a sequence $S\in \mathbb{C}$:The series has absolute convergence when ${\sum }_{n=1}^{\infty }|a_n|$ converges as a series of nonnegative real numbers. A series beginning at $n=0$ uses the same definitions with that starting index.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

The infimum of $A$ is its greatest lower bound: $\alpha =\inf A$ means that $\alpha \le x$ for every $x\in A$, and that every lower bound $l$ of $A$ satisfies $l\le \alpha$. The supremum is its least upper bound: $\beta =\sup A$ means that $x\le \beta$ for every $x\in A$, and every upper bound $u$ satisfies $\beta \le u$. The completeness of the real numbers ensures that both exist for a nonempty bounded set of real numbers.

Subtracting the first row from the other two in the determinant givesThis vanishes exactly when the two displacements $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ are linearly dependent. If $\mathbf{a}\ne \mathbf{b}$, that means $\mathbf{c}-\mathbf{a}=t(\mathbf{b}-\mathbf{a})$ for some real number $t$. Thus all three points lie on the lineConversely, putting three points on any one line makes their displacements linearly dependent, so the determinant vanishes.

The equivalence remains true without distinctness. If exactly two points coincide, the two distinct locations determine a line containing all three, and two rows of the original determinant coincide. If all three points coincide, any line through that point works and the determinant is again zero. Existence persists, although uniqueness of the line can fail.

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.

The extreme value theorem gives finite real numbersSince these values are attained, the set of strict upper bounds in the definition of $M$ is precisely $(\beta ,\infty )$, and the set of strict lower bounds in the definition of $m$ is precisely $(-\infty ,\alpha )$. They are nonempty and bounded on the required sides. By the definitions of infimum and supremum,

Eliminate the two pure second derivatives by taking $\alpha ,\beta$ to be the roots of $z^2+2Bz+C=0$. They are distinct real numbers exactly whenThus the required function is $H(B)=B^2$. The mixed coefficient becomes $4(C-B^2)\ne 0$, so the transformed equation is $y_{\xi \eta }=0$. Integrating once in each coordinate giveswith arbitrary sufficiently smooth functions $F,G$. This is the characteristic form of a strictly hyperbolic partial differential equation.

Source: released examination PDF, pages 180–184.

Solutions cover all seven questions, including every optional question. The module label is I, with M1 used for Mathematics and Computer Science; the formal paper code is A10138W1. Throughout Question 4, diagonalisable matrices are understood over the real numbers.

The three elementary row operations on an $m\times n$ real matrix are:

Each operation changes whole rows, so it acts on every column by the same rule. Each can be undone by an elementary row operation.

For a nonempty set $A\subseteq \mathbb{R}$, a supremum is a real number $s$ such that $a\le s$ for every $a\in A$, and every upper bound $u$ of $A$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $a\in A$ satisfies $a>s-\epsilon$. By completeness of the real numbers, it exists whenever $A$ is nonempty and bounded above.

An increasing monotone sequence satisfies $a_{n+1}\ge a_n$ for every $n$; a bounded sequence satisfies $|a_n|\le M$ for some finite $M$ and every $n$. Put $s=\sup \{a_n:n\ge 1\}$. Given $\epsilon >0$, the defining property of the supremum supplies $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesThis is exactly the definition of convergent sequence, so the bounded monotone sequence theorem gives $\boxed{a_n\to s}$.

For a nonzero solution, separation of variables gives $dy/y=kdx/(1+x^2)$, so $\ln |y|=k\arctan x+C$. Restoring the zero solution by allowing an arbitrary real multiplicative constant givesIt is valid for all real numbers $x$, including the constant solutions when $k=0$.

Use integration by parts with $u=\ln (1+x^2)$ and $dv=(1+3x^2)dx$, giving $v=x+x^3=x(1+x^2)$. The remaining integrand simplifies because $du=2xdx/(1+x^2)$:The real formula is valid for every real number $x$.

For a matrix proportional to its transpose, applying the transpose twice gives $A^T=5A$ and therefore $A=25A$. Over the real numbers or complex numbers, $24A=0$ implies

For a real number $x$, the hyperbolic functions satisfy $\cosh x>0$ and $1-{\tanh }^{2}x={\mathrm{sech}}^{2}x$. Hence $\cosh x=1/\sqrt{1-u^2}$, and multiplying it by $\tanh x=u$ givesThe sign of the numerator preserves the sign of $\sinh x$.

The complex modulus is a nonnegative real number. Equality therefore requires $z-1$ itself to be real and nonnegative. Its imaginary part gives $y=0$, and its real part then gives $x-1\ge 0$. Conversely every such point satisfies the equality. Hence the complete solution set isGeometrically this is the closed ray on the real axis starting at $z=1$.

All three linear equations are multiples of the first, leaving the single constraint $x+y+z=0$. Choose two real numbers $s,t$ freely:A basis for this two-dimensional linear subspace is $(1,0,-1{)}^T,(0,1,-1{)}^T$.

A common shift of all three angles leaves every spring length unchanged. This continuous rotational symmetry gives the zero mode: there is no restoring force against rigid rotation.

The equilateral equilibrium also has the dihedral group symmetry $D_3$. Its action on the three angular displacements preserves the uniform one-dimensional subspace and the two-dimensional sum-zero subspace. The latter is an irreducible representation over the real numbers; a reflection may reverse all angular signs as well as exchange bead labels, which does not change these invariant subspaces. The symmetric stiffness matrix must act as a scalar on the sum-zero subspace, explaining the equal vibrational angular frequencies. Directly, each sum-zero vector satisfies $(3I-11^T)\mathbf{v}=3\mathbf{v}$. This is an example of degenerate normal modes from permutation symmetry, and of normal modes of three equal masses with a symmetric gap potential.