Daily math problems for July 24, 2026 — Logic, Sequences, Algebra & more

Negation of $\mathrm{\forall }$ is $\mathrm{\exists }$; negation of $>$ is $\le$. So the negation is $\mathrm{\exists }x\in \mathbb{R}:x^2+1\le 0$. (This statement is itself false, but the question asked for the negation, not its truth value.)

$x^2=3\cdot 27=81\Rightarrow x=9$ (positive ratio).

$\bullet$ Simplify the constant term: $\sin \frac{\pi }{6}=\frac{1}{2}$, so $10\cdot \frac{1}{2}=5$:

$${25}^x-6\cdot 5^x+5=0$$

$\bullet$ Substitute $t=5^x>0$:

$$t^2-6t+5=0$$

$\bullet$ Factor:

$$(t-1)(t-5)=0\Rightarrow t\in \{1,5\}$$

$\bullet$ Back-substitute: $5^x=1\Rightarrow x=0$ and $5^x=5\Rightarrow x=1$.

$\bullet$ Compute the sum of squares:

$$x_1^2+x_2^2=0^2+1^2=1$$

$S_3$ (or equivalently $D_3$) has $6$ elements and is non-abelian. All groups of order $1,2,3,4,5$ are abelian.

$A={\displaystyle \frac{36\sqrt{3}}{4}}=9\sqrt{3}$.

$f^{\mathrm{\prime }}(x)<0\iff 0<x<2$. So $f$ is decreasing on $(0,2)$. (Increasing on $(-\mathrm{\infty },0)$ and $(2,+\mathrm{\infty })$.)

${\int }_0^{\pi \mathrm{/}2}{\sin }^{2}xdx={\int }_0^{\pi \mathrm{/}2}{\displaystyle \frac{1-\cos 2x}{2}}dx={\displaystyle \frac{1}{2}}{[x-\frac{\sin 2x}{2}]}_0^{\pi \mathrm{/}2}={\displaystyle \frac{\pi }{4}}$.

Euler's classical result: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}}$.

${\displaystyle \frac{p}{q}}$ runs over $\text{\hspace{0.17em}⁣}\{\pm 1,\pm {\displaystyle \frac{1}{2}}\}$.

$\bullet$ Simplify the exponent on the right: $i^{20}=(i^4{)}^5=1$, so $i^{20}+1=2$:

$$C_x^5=C_x^2$$

$\bullet$ Recall the combinatorial symmetry:

$$C_x^k=C_x^{x-k}$$

$\bullet$ Since $5\mathrm{\ne }2$, the symmetry forces $5+2=x$:

$$x=7$$