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 Jan26 comment Alternate proof for “$\log_{10}{2}$ is irrational” Oh, so that was your case... I've never seen the binary logarithm being written this way. lg yes, but log without the base specified explicitly is usually meant to be decimal by default where I live. I have this problem with Wolfram Alpha, too, which uses log for the natural logarithm instead of ln I used do. These disagreements on notation could be very confusing sometimes... Jan23 comment Alternate proof for “$\log_{10}{2}$ is irrational” Whoops, the original poster is you. Then are you going to contradict yourself now or something? I don't get it... Jan23 comment Alternate proof for “$\log_{10}{2}$ is irrational” @DavidRicherby: Blame the original poster, then, not me, since I just generalized his claim to show that $log 2$ isn't any more special than any other number being represented in a digital computer. Jan23 comment Proving/Disproving Product of two irrational number is irrational +1 for the simplest constructive proof, without diving into deep logical arguments of proof by contrapositive. Your way is how I'd do it myself: a constructive proof by counter-example. Of course it requires the prior knowledge that $sqrt{2}$ is irrational. But this is whole another story... Jan23 comment How to prove if log is rational/irrational You assume that $m$ and $n$ are integers. Then, when we get to the point that $2^m = 3^n$, you say that it says that even number = odd number, which is impossible. I agree. But what if we choose $a=0$ and $b=0$? ($0$ is an integer, right?). Then there's $0$ factors of $2$ on the left, and $0$ factors of $3$ on the right, that is, we get $1=1$, which is true, and your proof falls down. Am I right? Jan23 comment Alternate proof for “$\log_{10}{2}$ is irrational” @DavidRicherby: To a computer scientist, every number is rational, since this is how IEEE 754 standard for storing floating-point numbers works. It stores every number as a finite approximation by a binary fraction. There is no digital machine capable of storing an irrational number, and there never will. It is an inherent limitation of digital notation of numbers. Jan2 answered Relationship between Pi and Phi using the Great Pyramid of Giza? Dec19 comment Denesting Phi, Denesting Cube Roots How can mathematical formulas be copyrighted? If copyrighting math is possible, we are doomed. Dec18 comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ Yeah, about that... I've seen several of your answers, but I've nowhere seen the theory behind these two functions (norm and trace) explained. Is it elaborated somewhere a bit more? I can see how the norm formula is similar to how the norm of a complex number is calculated, but there are no imaginaries in nested radicals, so I don't quite understand WHY your magic works so well. And why is the norm being subtracted instead of divided out, and the trace quite contrary. Am I missing some obvious part of complex analysis or something? I'd like to know some background of how did you discover that. Dec18 comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ Sure, now I understand it. But let me show you what exactly was confusing for me at first: math.stackexchange.com/a/196181/16191 There you use the letter $w$ in calculating both the norm and the trace, so it seems as if they had to be calculated from the same number (the one under the original square root). Are you sure it doesn't need any correction? Then it will remain confusing. Dec18 comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ My second comment was about a separate problem. The first one was about the problem that when one calculates the trace from the original expression under the square root (as your other comments suggest), the result will be incorrect, because the trace has to be calculated from the intermediate result after the first step of your algorithm (as you correctly stated here in this answer, but incorrectly in others). That's what I meant that needs correction. Dec15 awarded Caucus Dec15 answered What is the square root of complex number i? Dec15 comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ Also, what if the norm turns out to be a square root too? Continuing the calculations lead to the answer 1/1 then, which is incorrect. (I see that it works that way when the expression under the square root is not a perfect square. But this has to be mentioned upfront: in which cases your algorithm does apply.) Dec15 comment How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ Can you correct it in your other answers too? I was a bit confused when I first stumbled upon one of them, since you put it in a way which suggested that one needs to divide out by the trace of the original expression under the square root, instead of the intermediate result. And it didn't work. Dec14 comment How to simplify a square root Oh, and one thing you wrote is a bit misleading: You put this in a way suggesting that one has to divide by the trace of the original square-rooted number ($w$), but when I tried this, it didn't work. One has to divide out by the trace of the partial answer obtained along the way instead, which is not the original $w$, but $w - \sqrt{norm}$. Dec14 comment How to simplify a square root Where does these names come from? ("norm" and "trace") I see that the first one is similar to calculating the norm of a complex number, but there's no $i$ in it, so why does it still work similarly to complex conjugates? And why is the norm subtracted instead of divided out? Why is it the trace which is divided out instead? The trick is really nice, but it would be nicer if you explained why does this magic work so well. Dec14 comment How to simplify a square root Nice technique and a good example of why there are really two solutions when taking the square root, contrary to what most mathematicians would like it to be (a single-valued function), and that it doesn't matter much whether you take the square root of an unknown or a "simple number". Dec14 comment How to simplify a square root But then you have to "guess" how to split that $27$. Dec14 comment How to simplify a square root What if the solutions to the quadratic were also irrational, so that $a$ and $b$ become nested inside their radicals again? Or is this possibility somehow excluded?