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 2d revised How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number added 32 characters in body 2d revised How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number added 108 characters in body 2d revised How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number deleted 9 characters in body 2d revised How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number added 37 characters in body 2d comment How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number and $-12\not\geq 0$... ! 2d answered How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number 2d revised Points on 3d line added 242 characters in body 2d revised Points on 3d line added 303 characters in body 2d answered Points on 3d line 2d comment What's special about the cauchy product? In your second paragraph the multiplicands $\sum a_n$ and $\sum b_n$ are completely independent, e.g. you can write $C_n=\left(\sum a_k\right)\left(\sum b_k\right)$. That's trivial. The Cauchy product combines terms in a non-trivial way. Cauchy products can be useful. Jul27 revised evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$ deleted 17 characters in body Jul27 revised evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$ deleted 41 characters in body Jul27 answered evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$ Jul23 revised Logarithmic Integral II deleted 251 characters in body Jul23 answered Logarithmic Integral II Jul23 revised How to solve $z^3 + \overline z = 0$ added 466 characters in body Jul23 comment Does the series $\sum\limits_{n=1}^\infty \frac{1}{n\sqrt[n]{n}}$ converge? +1 very simple ! Jul23 revised Simple Logarithms Equation added 30 characters in body Jul23 comment Simple Logarithms Equation Assuming $x\in\mathbb{R}$, and since $3^x>0$ for all $x\in\mathbb{R}$, then $3-x>0$ which implies $x<3$. Jul22 comment If I buy 2 lottery tickets do I double my chance of winning? Which gives $$\frac{1}{{49\choose 6}}=0.00000007151123842...$$ I'd better get a better paid job.