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December 27

real numbers - finite digits?

The exact decimal representation of Pi is a real number. If we remove the decimal point from the representation, i.e. yielding an 'integer-like' sequence of infinite digits with 3 as its most significant, etc, do we arrive at a real number? Since there is no 'last digit' we can also flip it - i.e. the least significant ("one's place") digit is 3, then ten's place is 1, and so on, utilizing the decimal expansion of pi but in the other direction.

Does this yield a real number?

(If the above question is unclear here is an equivalent: if we define a number as 1*10^0 + 1*10^1 + 1*10^2 etc - i.e. 11111 repeating - then is *this* number a real number?)

Why or why not? 212.96.61.236 (talk) 01:04, 27 December 2013 (UTC)[reply]

No it isn't. This is ruled out by the Archimedean property of the real numbers. Sławomir Biały (talk) 01:22, 27 December 2013 (UTC)[reply]

The Real number wiki page is not clear to newbies. Nowhere does it EXPLICITLY state that infinity is not a part of the Real Number system. 202.177.218.59 (talk) 02:20, 27 December 2013 (UTC)[reply]

The OP might be interested to read p-adic number which discusses a system very similar to that asked for. HTH, Robinh (talk) 08:26, 27 December 2013 (UTC)[reply]

Unfortunately, you have to click through to read about the Archimedean property. Sławomir Biały (talk) 17:19, 27 December 2013 (UTC)[reply]

(ec) First of all you're wrong in 'The exact decimal representation of Pi is a real number'. The decimal representation of Pi is not a real number, the decimal representation is a sequence of digits, and Pi itself is a real number. Next, the answer is 'no': the convention of decimal representation (and all other positional notations) does not assign any standard value to an infinite sequence of digits (exception: if a sequence contains only zeros from some place to infinity, you can drop those zeros and get a finite sequence of digits). --CiaPan (talk) 08:32, 27 December 2013 (UTC)[reply]

Using Complex Number in Scientific Calculators to calculate direction using Arg Button

Using scientific calculator to perform scouting compass navigation.

The complex number can be used to denote 2D mapping coordinates. And the direction from point A towards point B can be calculated using vector/complex number subtraction of "B - A" and then taking the Arg of the result.

However there is a problem. In maps, the magnetic compass angle starts from True North and increases in a clockwise direction. Where as in an Argand Diagram, the angle starts from "East" and increases in a counter-clockwise direction.

The solution is to have the imaginary unit pointing "East" and the (non-imaginary) unit pointing "North".

What that happens, you can use the scientific calculator to get the direction from point A to point B using the Arg button after calculating B - A. And the angle returned correspond to the angle on a magnetic scouting compass.

Have anyone in uses this in a scientific calculator? 202.177.218.59 (talk) 02:04, 27 December 2013 (UTC)[reply]

Is there a mathematical proofs on this (proofs)

Is there a mathematical proofs on this37.237.195.59 (talk) 16:51, 27 December 2013 (UTC)[reply]

(+x)(+y)=+xy

(-x)(+y)=-xy

(-x)(-y)=+xy

Formal proofs may depend on which definitions you use, which sets x and y belong to, and which axioms or proof methods you allow. If your math is below university level and they are real numbers or a subset then I suggest you just view them as "basically true by definition", especially the first. PrimeHunter (talk) 17:10, 27 December 2013 (UTC)[reply]

I don't think it's hard to suggest proofs that would satisfy pre-university level students. These are true because of commutativity and associativity of multiplication, and the fact that (+x) = (1)(x) and (-x)=(-1)(x). So a proof of the second one would be:

(-x)(+y) = (-1)(x)(1)(y) = (-1)(1)(xy) = (-1)(xy) = -xy

Proofs of the others would be similar. This assumes that you know how to multiply (-1)(1)=(-1) and other products of 1 and -1. If you want proofs for that, then I'd say it's "basically true by definition" of the multiplication operation. Staecker (talk) 00:53, 28 December 2013 (UTC)[reply]

If the goal is to satisfy students then their main issue is usually with (-x)×(-y) = x×y. They may not like (-1)×(-1) = 1 any better, or even -(-1) = 1. PrimeHunter (talk) 01:15, 28 December 2013 (UTC)[reply]

Note that -x is only a short hand for the additive inverse of x. The additive inverse is the number (element, in the more general case) which, added to x, results to zero. Let u be such that x + u = 0.

Writing -xy before we have proven that -(xy)=(-x)y=x(-y) is dangerous because it tempts us to use this in the course of the proof, so I will cautiously use lots of parenthesis.

Obviously u is the additive inverse of x, but also x is the additive inverse of u. So x=-u, u=-x and thus x=-(-x).

Now assume there was another number named v such that x+v=0, so x+u=x+v; now adding u you get u+x+v=u+x+u, and from that v=u. This proves the additive inverse is unique.

Multiply u+x=0 with y, and you get uy+xy=0. So uy is the additive inverse of xy, in other words (-x)y=-(xy). Do the same with x and y exchanged, you get x(-y)=-(xy). Do the same with -x instead of x and you get (-x)(-y)=-((-x)y). -((-x)y) is the additive inverse of (-x)y which in turn is the additive inverse of xy, thus (-x)(-y)=xy

95.115.187.53 (talk) 14:21, 29 December 2013 (UTC)[reply]