Mathematics!!

[GoldLynx]

Glad I had a question that got stuck in some ones head lol

 

Thanks for the answers though, that does makes sense. Prime is a property of the number regardless of what symbols represent it. Can finally get this one out of my brain.

 

Although Gainsdalf's comment up there about "symbols used to represent the number, not the number itself" leads down some almost metaphysical rabbit hole. Although I suppose it must be true that numbers and the symbols that represent them are separate things.

Loved this question/discussion. Simply top-notch. Prime.

[GoldLynx]

Wow.  So I'm not even close to on par with all y'all.  I don't understand half the crap in this thread.  

 

That said, I do like math that applies to things.  I have a B.S. in Automotive and quite enjoy figuring numbers as they relate to mechanical operation of engines and things.  Finding a way to cram more fuel/air into an engine and bump up the torque/horsepower numbers is always fun.  

 

Course, it's not quite as much fun as using the extra power to shred some tires.......

I am jealous of your skillz. My car knowledge is extremely looooowww. Kudos to you, sir.

[GoldLynx]

Wouldn't the inside angle of a corner of an equilateral triangle be 1/6 of a circle? And that of an octagon would be 3/8. Not sure I'm tracking the math.

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Those are the numbers I got, if I understood the concept correctly. Yay geometry!!! Im not sure what i think of 360° being considered 1. Ithink it would slow me down.

[Sam Ashen]

Wouldn't the inside angle of a corner of an equilateral triangle be 1/6 of a circle? And that of an octagon would be 3/8. Not sure I'm tracking the math.

 

It looks like the outside angle was intended.  So if I was walking along an equilateral triangle, I might walk 100 meters due north; turn right 120 degrees and walk another 100 meters; then turn right another 120 degrees and walk another 100 meters.

 

I am with Wovercast on Radians.  If I would get around to looking up the ASCII codes for the Greek letters, then I have a shot at taking a more active part in these discussions.

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Here's one for the crowd:

 

What do you do for terms in brackets that run on past the page?

 

Say z = a(bcd+efg+hij+klm+nop+qrs+tuv+wx)y

 

I usually stack the terms in a big square bracket when I scribble it on a piece of scrap paper, but I have no recollection of how we were taught to do it back in fifth grade.  So ya.  There are plenty of fifth-graders out there who are smarter than I am!  

[AugustaAdaByron]

If I would get around to looking up the ASCII codes for the Greek letters, then I have a shot at taking a more active part in these discussions.

 

If it helps I can write the greek alphabet and you can copy paste.

 

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο Ï€ Ï Ïƒ Ï„ Ï… φ χ ψ ω

 

Α Î’ Γ Δ Ε Ζ Η Θ Ι Κ Λ Îœ ΠΞ Ο Π Ρ Σ Τ Î¥ Φ Χ Ψ Ω 

[Sam Ashen]

If it helps I can write the greek alphabet and you can copy paste.

 

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο Ï€ Ï Ïƒ Ï„ Ï… φ χ ψ ω

 

Α Î’ Γ Δ Ε Ζ Η Θ Ι Κ Λ Îœ ΠΞ Ο Π Ρ Σ Τ Î¥ Φ Χ Ψ Ω 

 

All Bigots Get Diarahea Eventually

Zorro Ate The Ice Cap

Let's Munch Nuts eXcessively

Okay Pigs Really Stink Terribly!

If this kind of sounds crazy

Under Phive Chairs, Psichiatrists Wink.

[AugustaAdaByron]

All Bigots Get Diarahea Eventually

Zorro Ate The Ice Cap

Let's Munch Nuts eXcessively

Okay Pigs Really Stink Terribly!

If this kind of sounds crazy

Under Phive Chairs, Psichiatrists Wink.

 

It took me a while to decode.

[caeliter]

Since the primes question was already answered, here's something cool my dad and I realized that changing the counting system to a different base DOES do.

 

For any numbering system base "n" the sum of the digits of any multiple of n-1 will be a multiple of n-1.

 

To use our standard base 10 counting system, 9, 18, 27, etc. To use a base 8 system (cuz... why not? first number that occurred to me) 7, 16, 25, 34, etc. and going the other way, base... let's say 12? B, 1A, 29, 38, etc.

 

A couple follow ups that I'm pretty sure of, but have not verified:

 

If you repeat this process with the results, eventually you will reach n-1 (example, 999 ->27 -> 9)

 

For any number that is a factor of n-1 the sum of it's digits will add up to a multiple of that factor. (in base 10, we know this to be true of 3; using a base 13 system because I wanted a nice even number for n-1, 13 is 4 squared, 20 is 2*D, 3*3*B = 78 -> 12 ->3)

 

It's one of those things that totally makes sense when you think about it, but why would you ever think about it?

[AugustaAdaByron]

More generally, what's true is the following:

 

For any base b and any number n represented in base b. If you repeatedly add its digits until you end up to a single number you get the remainder from the division of n by (b-1) in the base system b (or b-1 if n is a multiple of n-1).

[Artinum]

BSc Mathemathics here, but I graduated fifteen years ago (!!) and I may have forgotten most of it by now.

But I do have an innate understanding of how computers think.

[caeliter]

More generally, what's true is the following:

 

For any base b and any number n represented in base b. If you repeatedly add its digits until you end up to a single number you get the remainder from the division of n by (b-1) in the base system b (or b-1 if n is a multiple of n-1).

 

For some reason my brain is having a hard time with the wording of that...

I'm going to take a couple examples and see if I'm understanding correctly

 

So if I convert 3498726 to base 14 I get 671090 who's digits add up to 19 (in base 14) who's digits add up to A. 3498726/13 does indeed produce a remainder of 10.

Ok, this is cool, lets try one more for good measure...

5419821 converted to base 5 is 2341413241 -> 100 -> 1; 5419821/4 gets remainder 1.

 

Oh snaps! Ok, I don't know why I was having trouble parsing that at first, but it totally makes sense now that I've done a couple.

[AugustaAdaByron]

For some reason my brain is having a hard time with the wording of that...

I'm going to take a couple examples and see if I'm understanding correctly

 

So if I convert 3498726 to base 14 I get 671090 who's digits add up to 19 (in base 14) who's digits add up to A. 3498726/13 does indeed produce a remainder of 10.

Ok, this is cool, lets try one more for good measure...

5419821 converted to base 5 is 2341413241 -> 100 -> 1; 5419821/4 gets remainder 1.

 

Oh snaps! Ok, I don't know why I was having trouble parsing that at first, but it totally makes sense now that I've done a couple.

 

Oops, I should have given some examples. Sorry!

 

The reason this happens is this. Take for example the number 5419821 converted to base 5 that you use.  

Then this number in base 5 is actually:

 

1x5⁰ + 4x5¹+2x5²+3x5³+1x5⁴+4x5⁵+ 1x5⁶ + 4x5⁷ + 3x5⁸ + 2x5⁹.

 

But you can write 5 as 4+1 and rewrite the above as: 

 

1x(4+1)⁰ + 4x(4+1)¹+2x(4+1)²+3x(4+1)³+1x(4+1)⁴+4x(4+1)⁵+ 1x(4+1)⁶ + 4x(4+1)⁷ + 3x(4+1)⁸ + 2x(4+1)⁹.

 

Using the binomial expansion we obtain that each one of the (4+1)^z is of the form 4f(z) +1. Thus, the number becomes the sum of its digits plus something that is divisible by 4. (I skipped rewriting. The reason is that it really get really non-parsable if I expand it more).

 

Using a similar method you can try to see what happens when instead of dividing by b-1 you divide by b+1.

 

 

 

[caeliter]

I enjoy math, but am a bit unpracticed so I'm going to try and work my way through it.

 

Using the same example:

 

instead of 4+1, we're then looking at 6-1 so: 

 

1x(6-1)⁰ + 4x(6-1)¹+2x(6-1)²+3x(6-1)³+1x(6-1)⁴+4x(6-1)⁵+ 1x(6-1)⁶ + 4x(6-1)⁷ + 3x(6-1)⁸ + 2x(6-1)⁹

 

So we're looking at each term alternating between 6f(z) + 1 and 6f(z) -1

 

so alternating adding and subtracting should give us the remainder when dividing by 6? (If my logic hasn't broken yet)

 

so 1-4+2-3+1-4+1-4+3-2 = 8 - 17 = -9 (oops forgot to put it in base 5 >.<) -14 -> 3

 

Ok that totally worked. Though I'm assuming it's possible to get a negative value (so that value + b+1 will get the remainder)

[AugustaAdaByron]

That's correct!

[Oramac]

My brain hurts now......

[Estrix]

How about a more science question?

 

A few years ago it was announced (in error as it turned out) that the LHC had detected faster than light particles. In breach of the rules of special relativity (I think). However, if the particle in question is born/arises at a super-luminary speed would that in fact breach special relativity?

[goodbye_farewell]

How about a more science question?

 

A few years ago it was announced (in error as it turned out) that the LHC had detected faster than light particles. In breach of the rules of special relativity (I think). However, if the particle in question is born/arises at a super-luminary speed would that in fact breach special relativity?

In short, no. Special relativity allows for particles to travel faster than the speed of light as long as they are always traveling faster than the speed of light.

[AugustaAdaByron]

I have no background in physics and I'm also not sure if what I'll say is the same with what wovercast said. Let me try though.

Any massless particle that travels faster than the speed of light does not breach the rules of special relativity. Unless I am mistaken, if they discover a massive particle (that is, with non-zero rest mass) that travels faster than the speed of light then that would breach the rules of special relativity.

[goodbye_farewell]

I have no background in physics and I'm also not sure if what I'll say is the same with what wovercast said. Let me try though.

Any massless particle that travels faster than the speed of light does not breach the rules of special relativity. Unless I am mistaken, if they discover a massive particle (that is, with non-zero rest mass) that travels faster than the speed of light then that would breach the rules of special relativity.

 

Keep in mind that this is all theoretical, but no, particles of finite mass can have velocities greater than c. However, they cannot move from below c to greater than c. They have to ALWAYS be above c. In fact, relativity prohibits it from slowing down to under c.

[AugustaAdaByron]

Keep in mind that this is all theoretical, but no, particles of finite mass can have velocities greater than c. However, they cannot move from below c to greater than c. They have to ALWAYS be above c. In fact, relativity prohibits it from slowing down to under c.

 

That makes sense. That's because of the formula for the relativistic mass, right?

[goodbye_farewell]

That makes sense. That's because of the formula for the relativistic mass, right?

You got it. Imaginary mass is a funny thing...

[Shen]

Wow, what a great thread!! I went through Euclid's elements, Ptolemy, Copernicus, and Kepler, Descartes, Newton, Leibniz, Einstein, and all the way up through non-Euclidean geometry in undergrad. Now I'm working on a masters in conservation biology and am trying to teach myself R programming, statistics, and trying to re-learn calculus (which is tricky without all the pesky algebra stuff I learned and promptly forgot for 10 years). I love R, though!

[Artinum]

You got it. Imaginary mass is a funny thing...

 

Isn't that what lapsed Catholics have? :-P

Sorry, this isn't the bad joke thread, is it?

 

I don't know about imaginary mass, but I do like complex numbers. I've written Mandelbrot generators in several programming languages.

[AugustaAdaByron]

You got it. Imaginary mass is a funny thing...

 

It wouldn't even get the chance to become imaginary. It'd get divided by zero first and then...

 

[Gainsdalf the Whey]

So I understand the concept of something beginning at c never being able to accelerate above it as it takes infinite energy to accelerate to c in the first place, what's the math behind the faster than light particles not being able to deccelerate below it?