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For fixed $m,l$, setAt $n=0$, $D_0=F_l{F}_m$. Applying the Fibonacci recurrence to every term and cancelling gives $D_{n+1}=-D_n$. Induction therefore proves

Applying this identity to the relevant pairs of indices and using $F_{r+1}=F_r+F_{r-1}$ givesandThus, for $j\ge k\ge 0$,and

For $n=1$,If $C_n=F_{n+1}{F}_{n-1}-F_n^2$, then the recurrence givesInduction yields the Fibonacci determinant identity known as Cassini's identity:

LetBy Fermat's little theorem, $l\mid p-1$. If the digit count $L$ is a multiple of $l$, then $10^L\equiv 1(\mathrm{mod}p)$. The cyclic decimal divisibility relation givesand hence also modulo $p$. Since $p>7$ is coprime to $10$, every $10^j$ is a unit modulo $p$. Thereforefor every $j$. Since $c_0(n)=n$, this proves the required equivalence.

As printed, the claim is false: $70$ is 7-cyclic-divisible because its rotations are $70$ and $07=7$, but its digits are not all equal to $7$ and its length is not a multiple of $6$.

The standard rotation argument proves the likely intended statement. If an $L$-digit integer and all its rotations are divisible by $7$, then for each leading digit $a$,is divisible by $7$. The multiplicative order of $10$ modulo $7$ is $6$. If $6\nmid L$, then $7\nmid 10^L-1$, forcing every digit $a$ to be divisible by $7$, hence to be either $0$ or $7$. Thus the valid conclusion is:The counterexample shows why “all its digits are equal to 7” cannot replace the first alternative.

Divisibility by $2$ or $5$ depends only on the final digit. As the digits are cyclically rotated, every digit appears in the final position. Thus every rotation is divisible by $d\in \{2,5\}$ exactly when every digit is divisible by $d$.

A decimal integer is congruent modulo $3$ to its digit sum. Cyclic permutation does not change that sum. Hence every rotation of a multiple of $3$ is again divisible by $3$.

Again set $c=1$ and writeA boost of velocity $v$ givesDefineThe hypothesis says $D(0)<0$. Put $\xi =x_B-x_A$ and $\tau =t_B-t_A$. Causality gives $\tau \ge |\xi |$, while $T_B\ge |X_B|$, and $T_A=T_B+\tau$, $X_A=X_B+\xi$. ThereforeButHence at least one of $D(1),D(-1)$ is positive. By continuity, $D(v)>0$ for some physical velocity $|v|<1$ sufficiently close to that endpoint. In the corresponding inertial frame,as required.

For a lamina in the $xy$-plane,so the perpendicular axis theorem is immediately

For the uniform ellipse, put $x=au$, $y=bv$, where $u^2+v^2\le 1$. Symmetry of the unit disk givesand the same for $v^2$. Hence

The parallel axis theorem says that for an axis through $x$, parallel to the centre-of-mass axis in direction $\hat{d}$,LetMoments about the axis through $x_C$ add, soUsing $M_A=M_C-M_B$ and solving giveswhen the denominator is nonzero. In the degenerate equal-distance case, this inertia equation alone does not determine $M_B$; the centre-of-mass equation $M_C{x}_C=M_A{x}_A+M_B{x}_B$ supplies the remaining information.

This relation is reflexive and transitive, but not symmetric: $1\sim 0$ whereas $0\nsim 1$. It is therefore not an equivalence relation.

This is an equivalence relation: $x-x=0\in \mathbb{Z}$; if $x-y\in \mathbb{Z}$, then $y-x\in \mathbb{Z}$; and integer differences add. Its classes arethe cosets of $\mathbb{Z}$ in $\mathbb{R}$.

The relation is reflexive and symmetric, but not transitive. For example,while $1\nsim -1$. Hence it is not an equivalence relation.

This is not an equivalence relation because it is not reflexive: $0\nsim 0$, since $0\cdot 0\ngtr 0$.

Momentum conservation is $P_A+P_B=P_C+P_D$. The diagonal bilinears are $P_a^2=m_a^2{c}^2$. The six off-diagonal ones areandThese express every bilinear invariant through the Mandelstam variables and masses. Expanding $s+t+u$ and using momentum conservation cancels all mixed products, leaving