caledonia

I seek a proof, (using the properties of the h.c.f.), that if k divides nr then k = k1k2 where k1 divides n and k2 divides r

Gauss periods, used to solve the heptadecagon, are defined in such a way that each level can be determined from the one above. For example Gauss periods of length 4 determined from those of length 8. Can anyone suggest how Gauss came to this conclusion. Or offer a proof ? Thanks

I like to imagine two masses A and B, far enough from all other bodies that the gravitational attraction of the latter is negligible. Newton gives us a formula for calculating the attractive force between A and B. Is the formula exact, even for very large / small masses, and for a very large / small distance between A and B ? OR, does GR kick in and predict discrepancies even in this simplest of scenarios ?

"We start seeing deviations...". My question concerns these deviations. I asked how the GR force is calculated, and if it is greater or less than the Newtonian formula predicts.

Newton gives a formula for calculating a gravitational force. Is there a similar formula for GR ? Is the einsteinian force (always) greater than, or less than, the newtonian ?

Sorry. I accept that my posting should have been a blog. I would therefore like to remove my original posting (preaching?) but do not know how to do that. Feel free, if anyone does know . . .

If thou wouldst follow Newton in his Book 1, Section 3, Proposition XI and deduce the inverse square law from Kepler's observations of elliptical planetary orbits, first know thy Conics. In particular, the proof uses these properties of the ellipse : 1. The sum of the lines joining the foci to any point on the ellipse is constant 2. The lines joining the foci to a point on the ellipse make equal angles with the tangent there 3. The area of the parallelogram formed by the tangents at the ends of any diameter is constant 4 . The 'symptom' of the ellipse viz. for any diameter, the ratio of the square of an ordinate to the product of the abscissae is constant Newton and his contemporary mathematicians were well versed in Apollonius where these and many other properties of the Conics are proved. But Apollonius is long and difficult : fortunately we have shorter methods : Dandelin spheres easily prove 1. It is easy to deduce 2 from 1. Archimedes proved in Proposition 9 of his "Spheroids and Conoids" that we may regard the ellipse as a slant section of a right cylinder rather than a cone. Then for any diameter, the orthogonal projection of the parallelogram in 3 is a square describing the circular 'base'. All such squares are of course equal in area. Again using the cylindrical definition of the ellipse, the abscissae / ordinate in 4 are projected onto a diameter and perpendicular ordinate of the circle. It is easy to prove the constancy of the ratio of (the square of) an ordinate and (the product of) its abscissae for the circle. Now goest thou and follow Newton . . .

I too am still using Windows XP. I tried Windows 10 on my PC and could not find any advantages - but plenty of pain! It may be a swell operating system for touch-screen tablets and smart phones. Why are you finding XP "increasingly difficult to use" ? If you are forced to switch for some tasks, I suggest that you keep your XP computer operational and have a new Windows 10 device running alongside.

Interesting repies - thank you. But none offer what I ask for viz. a minimal scenario where Time Dilation operates, but Doppler does not. As I suggested, it may be impossible to have such a situation.

When we are stationary relative to each other, I see a (very) distant beacon flashing once per second. Is it correct please that if I move towards this source at high speed (say 1/2c) I perceive the flashes to be faster ; and that this a consequence of both TD and RD ? Conversely, moving away the beacon appears to slow down ? I would like then to envisage a scenario (gedankexperiment) where TD only comes into play - if this is possible ? Thank you

The atomic bomb demonstrates the conversion of mass-into-energy. Are there known (cosmological) phenomena which are instances of energy-into-matter ?

Thanks for responses. I was thinking about a proof which did not require the explicit identification of a root, but I accept that cos 2pi/n + i sin 2pi/n is indeed a primitive root (of xn = 1).

I seek a proof that xn -1 = 0 has a primitive root.

Bad spacing in above post, better now with monospaced font : I think I have a direct arithmetical proof : To take a numerical example : to prove that 1.2.3...30 divides any 'block' of 30 bigger numbers e.g. 53.54.55...82. We think of 'cancelling' primes from 'top and bottom'. Consider the occurence of the prime 3 in the bottom - here is the distribution : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 1 2 1 1 2 1 1 3 1 14 altogether and for the top : 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 3 1 1 2 1 1 2 1 1 4 17 altogether The highest power of 3 present in the bottom is 3, viz 3 cubed = 27 occuring once. There must be at least one multiple of 27 in top (in fact there are two), so we can divide it by 27 and cancel 27 from the bottom. Next, there remain 2 occurences of 3 squared among the bottom numbers, at 9 and 18. Similarily there must be multiples of these numbers on the top - we see them at 63 and 72. Divide 63 and 72 by 9 and 18 to cancel these 3s. Continuing in this way with remaining single powers of 3, we can cancel them all from the bottom (three will be left on top). The position and number of other primes present in the bottom are unaffected so far. Hence we can pick any other prime and repeat the process. Eventually all numbers will be gone from the bottom ! QED

I think I have a direct arithmetical proof : To take a numerical example : to prove that 1.2.3...30 divides any 'block' of 30 bigger numbers e.g. 53.54.55...82. We think of 'cancelling' primes from 'top and bottom'. Consider the occurence of the prime 3 in the bottom - here is the distribution : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 1 2 1 1 2 1 1 3 1 14 altogether and for the top : 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 3 1 1 2 1 1 2 1 1 4 17 altogether The highest power of 3 present in the bottom is 3, viz 3 cubed = 27 occuring once. There must be at least one multiple of 27 in top (in fact there are two), so we can divide it by 27 and cancel 27 from the bottom. Next, there remain 2 occurences of 3 squared among the bottom numbers, at 9 and 18. Similarily there must be multiples of these numbers on the top - we see them at 63 and 72. Divide 63 and 72 by 9 and 18 to cancel these 3s. Continuing in this way with remaining single powers of 3, we can cancel them all from the bottom (three will be left on top). The position and number of other primes present in the bottom are unaffected so far. Hence we can pick any other prime and repeat the process. Eventually all numbers will be gone from the bottom ! QED