daniel_haxby

If you're refusing to graph because you cannot find a straight edge to hand... I know it is trivial. Buuuuut, the guy who posted the questions had obviously only half understood what is a valid technique for solving modulus problems, and while in this case it is unnecessary/pointless, for the future might be worth clarifying. That was all I was trying to explain. Yes, I did not explain that bit correctly. I didn't go back and read the posts properly. PS and I wouldn't graph it either, but the original questions were Solve and Graph the following.

I have seen the video. I wasn't aware we were on a serious thread. Read irony.

I would disagree that the border between the two is artificial, alan2. Discrete maths is not a simplification of continuous problems (though it can be used as such, e.g. computational methods). The study of discrete maths like graph theory, combinatorics, number theory, finite group theory, etc. are all studied for their own sake, and owe continuous maths no favours or provenance. The fact that sometimes discrete maths comes up in the study of continuous problems just goes to show that the world works in more discrete ways than we think!

I do actually know there is no solution. But if you had to show that graphically, you would graph y=f(x), and look for points y=x. If you do graph the two lines I gave, in their respective domains, then you get a graph that is a horizontal line all above the y axis, so there is no solution. I know it is futile, but the method isn't, just the question.

Guys, it's a sea-launch platform. The satellite launches from the sea, not the platform! Duh!

The money will be in computer science, which needs discrete maths, so if that's how you want to measure it. Or perhaps because of computers the time for advances in discrete maths is at hand. I believe (and if my supervisor is to be believed) that a lot of cutting edge physics topology is described by finite group and ring theory. I very much prefer finite algebra type stuff, so any future maths I do is not going to be continuous!

[Not an expert] But as you say, an irrational number alone is too predictable. Any transposition system (ie, encoding) that does not use a CSPRNG to encode it will itself be insecure, and using one makes the irrational number as a CSPRNG redundant. Any larger method you are describing doesn't seem to include orders of magnitude more complexity, and even if it did, you are keying the encoding with something not a suitable CSPRNG, so again it will be weak unless you use unfeasibly large numbers of keys (irrational numbers).

I mean no x=y, hence no solution.

But you can graph the two equations I gave, and point out that no x lies in the image of f(x).

When I started writing my reply (#53) we were on post #46! And I think I may have attempted to derive Compton time, as Jaques mentioned in #46! This is me learning physics.

This debate is interesting me in both TurricaN's predicament and Martin's knowledge of Planck units (measurement?), so here is my attempt to put a different angle on things. I may be incorrect, so feel free to post corrections; that's why I'm posting. On the other hand, I may be writing exactly what TurricaN thinks he wants, in which case, it may illustrate Martin's point. Start with [math]E=mc^2[/math] and [math]E=vh[/math], where v is frequency of a photon, and m is that photon's mass, c is light in a vacuum, and h is Planck constant (not h-bar yet). So, equate to get [math]vh=mc^2[/math]. So, from what I believe Martin was saying, for this to make sense in the universal case, we need to use Planck units, or rather consider the thought experiments from which they are derived. We have figures for converting the Planck units to SI, so by using these in the correct manner we should get some conversion into SI. For v, we mean [math]v=v_0 /t_{\textrm{unit}}[/math], where [math]v_0[/math] is a quantity, and [math]t_{\textrm{unit}}[/math] is the unit we are measuring in, which is Planck time. For universality, we pick 1 Planck time. Now we convert Planck time to seconds by multiplying t(unit) by the conversion factor to SI (seconds), which is [math]t_p[/math]. Now we have [math]h v_0 / t_p = mc^2[/math] For m to make sense we must consider the photon whose frequency we were measuring, and its mass is [math]m=m_0 m_{\textrm{unit}}[/math] where m_0 is a quantity, and m(unit) is 1 Planck mass. To get kg, we use conversion factor m_p to say [math]m=m_0 m_p[/math] Now, [math]h v_0 / t_p = m_0 m_p c^2[/math] But [math]t_p = \bar{h} / m_p c^2[/math], So [math]h v_0 / t_p = 2 \pi \bar{h} v_0 / t_p = 2 \pi \bar{h} v_0 c^2 m_p / \bar{h} = m_0 m_p c^2[/math] Which means [math]v_0 2 \pi = m_0[/math] which I presume means that a photon of mass m Planck units takes its own mass divided by 2pi units of Planck time to cross its own wavelength. Comments or corrections to working or interpretation of result welcomed. To TurricaN, I hope that final equation demonstrates what Martin was saying, that you can only equate time with mass in a meaningful way in the thought experiment in which Planck units are derived (defined?). You can't convert time into mass, but you can relate the frequency of a photon with its mass.

This might help: http://www.cut-the-knot.org/do_you_know/compass.shtml The question you ask is not immediately addressed by the constructions given, but the website assures the reader the proof/method you seek is available as a combination of those constructions given. I didn't have the patience to check if this is the case! But it looks like what you're after.

(Disclaimer: pop science/non-expert answer!) It makes it a bit difficult to have faith in a deterministic universe. Even if everyone has their eyes open, a vast amount happens far away in the universe that we cannot observe. You may suppose this doesn't affect us, since to affect us, we'd observe its effect, and then we wouldn't have a deterministic quandry. But consider a closed box, weighing mass M, and opaque to every force we can detect. Imagine if it opened. What would come out? We start observing the interior of the box, and so this mess of quantum superposition breaks down to some definite answer. But why should the answer it gives make any sense at all? Anything could be in there, and not just anything we could predict or conceive of. In an objective universe, when a photon comes flying out of the distances of the universe, we learn something about possible future photons. But if the universe can do whatever it wants to when we're not looking, what's the point?

Here, you've done everything correctly to this point, except not noticed that you've inexplicably swapped the inequality sign around. If you did that on purpose, you've made a mistake. So go back to check the question. If the question you gave us is correct, then there is no d that satisfies the equation. You can still graph y=x-9/4 and y=x-5/2 to demonstrate graphically that there are no x above the first line which are below the second line. If the question you gave us is incorrect, and the inequality faces the other way, then you got to 12d-27>12d-30. From there you can add 27 to each side, and then you should be able see the answer, and make a graph, although, like last time, you don't need to. I know what is meant by two lines here. A technique taught, which I think is silly, but some people like, is to make two equations which are the equations corresponding to: |x|=x ; for x positive |x|=-x ; for x negative I think that while it is a good method for graphing, it is incredibly confusing unless you understand what absolute value represents, and then why would you need to be told this technique?! The two lines you want are actually: 4x-3=-27 ; for x positive -4x-3=-27 ; for x negative You see, you only change the sign of the x in each case, nothing else. Now you can graph these two lines, using each graph only in its respective domain. An equivalent method is to draw both lines in full, but delete anything falling below the x axis. Wikipedia 'absolute value' to see representative graphs. Now, how do you graph these two lines in particular? The only 'meaningful' way will be to draw x=k, x=-k, delete the bits inappropriate, and then find where x crosses the y axis (!).

The babylonians had a complex but comparatively sophisticated (for its time) method of division, which used lots of factors and sums of fractions. They used base 60 notation because 60 has a lot of factors, so it gives you a lot of fractions. While we cannot be sure why they settled on base 60, since there is no historical evidence for divine or demonic instruction to their engineers, scribes or mathematiciens, it would be simpler to assume their choice was pragmatic in view of their arithmetical tools. If you have any sources to back up your assertion, being interested in the history of maths, I'd love to see them