I can see that $\sqrt{1/2} = 1/\sqrt{2}$ My calculator also confirms I can change the denominator and the equality still holds. But $\sqrt{2/3} \neq 2/\sqrt{3}$ Can someone explain why? I need to get the concept here.

I need to get from $[(1-p)f+p(1-f)](1+v)-[(1-p)(1-f)+pf] = x$ to $(2+v)(f+p-2pf)-1 = x$ but I’m stuck. I’d appreciate any tips on what I should I do after the following. $(f+p-2pf)(1+v) + (f + p – 2pf) – 1 = x$ Thanks in advance.

If the following quadratic equation $$qx^2+(p+q)x+bp=0$$ always has rational roots for any non-zero integers $p$ and $q$ what will be the value of $b$? My book’s solution says the value of $b$ will be $0$ or $1$. If we consider the discriminant of the equation, $$D=(p+q)^2-4bqp = p^2+2q(1-2b)p+q^2$$ then $D$ should be a perfect square […]

My question is: Solve: $|x-4|< a$, where $a$ belongs to the real numbers. Solve this by considering various cases depending upon whether $a$ is negative, positive or zero. What I have tried so far: If $a>0$ then: $x < a+4$ and $x>4-a$, if $a=0$ then there is no solution. My doubt is: Should I consider […]

If $a,b,c \in R$, then prove that: $$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b} \leq \frac{a+b+c}{2}$$ I can’t see any known inequality working here like $H.M.-A.M.$. Could this be solved using basic inequalities?

Alright, I need to find the partial fractions for the expression above. I have tried writing this as $$\frac{a}{x-1/2}+\frac{b}{x+1/3}$$ but the results give me $a=25.8$ and $b=-20.8$, which are slightly wrong because they give me $5x+19$ instead of $-5x+19$. Can you please help? Thanks a lot

This is a pretty dumb question, but it’s been a while since I had to do math like this and it’s escaping me at the moment (actually, I’m not sure I ever knew how to do this. I remember the basic trigonometric identities, but not anything like this). I have a simple equation of one […]

I have following equation to solve for $x$ $$\ln\left(1+\frac{bx}{a}\right)=\frac{4cx}{a}$$ where $a>0,b>0$ and $c>0$. In my own attempt I replaced $1+\frac{bx}{a}$ by $y$ and with this replacement the final form of the equation is $$ye^{-\frac{4cy}{b}}=e^{-\frac{4c}{b}}$$ I don’t know how to proceed further. Any help in this regard will be much appreciated. BR Frank

How to show this function is increasing convex function: Define $f(z)=\frac{T(z)}{g(z)}$, where $T(z)=\phi(z)-\alpha \phi(\frac{z}{\alpha})+z(\Phi(z)-\Phi(\frac{z}{\alpha}))\,,$ $g(z)=\Phi(z)-\frac{1}{2}\Phi(\frac{z}{\alpha})-\int_{-\infty}^{z}\phi(x)\Phi(\sqrt{\frac{1-\alpha^2}{\alpha^2}}\,x)) dx$ and $\Phi(z)=\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ is the CDF of standard normal distribution and $\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$ is the PDF of standard normal distribution and $0<\alpha<1$. How can we show $f(z)$ is increasing and convex (numerical graph of the function is increasing and convex)? (maybe […]

Prove the conjecture or give a counter-example: For each $m\in\mathbb N$ there exist a $n>m+1$ such that $m^2+n^2+(mn)^2$ is a perfect square. I have just tried it out numerically and it holds for $m<1000$. I can’t see any pattern for the smallest $n$: m n 1 12 2 8 3 18 4 32 5 50 […]

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