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z9u2K

bannanas and apples...

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Okay, You''ll probably think I''m an idiot for asking this, but I was wonderng if someone has a proof to the fact that the result of adding/subtracting/multipling/dividing two real numbers is a real number and not an imaginey number... I know, I know... You''ll all gonna say "Well duh! you take a banna and a banna and you think you will get an apple?!" But it doesn''t work that way in math... You gotta have a proof for everything you use... right? Is there a proof for that or it is an axiom?

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From what I know you can only get an imaginary number from Rooting a negitive number by something. It has been a while though.

For example:
Sqrt(-4)
i Sqrt (4)
2i

[edited by - Prod on August 12, 2002 1:49:28 PM]

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I don't know if this is convincing enough, but any real number is a complex number whose imaginary part is 0:

(5 + 0i) + (9 + 0i) = (14 + 0i) = 14

What do you need the proof for?

Cédric

[edited by - cedricl on August 12, 2002 1:52:52 PM]

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Consider the following definition given by www.m-w.com:
Main Entry: real number
Function: noun
Date: 1909
: one of the numbers that have no imaginary parts and comprise the rationals and the irrationals

Let us use i = (-1)^1/2 (the square root of -1).

Usually, a COMPLEX number is written as A + Bi where A is the REAL part, and B is the COMPLEX part of the number.

In a REAL number, the COMPLEX part is 0, thus our number, Z, can be written as:

Z = A + 0i = A

Now then, say we ADD two real numbers, Y and X where

X = D + 0i
Y = E + 0i

Then

X + Y = (D + 0i) + (E +0i)

We group like coefficients (1 and i) to get

X + Y = (D + E)*1 + (0 + 0)*i

We see that our imaginary part is ZERO, regardless of what the values for D and E are -- since any number times zero is zero.

Likewise, if we MULTIPLY two numbers X*Y:

X*Y = (D + 0i)(E + 0i)

Expanding, we get:

X*Y = D*E + 0*E*i +0*D*i + 0*0*i*i

Again, this is equal to D*E because all other terms are multiplied by zero.

Perhaps this is not as formal a proof as you would like, but it should set you on the right path.

Best of luck,
Steve

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If you are considering the real numbers as a field on their own, rather than extended to the complex field, then I think that the operations like + and * are only defined on that field. That is to say that it is only possible to add two numbers in the field, and the result must be in the field. So two real numbers add to produce another real number (or any other such arithmetic operation I think). So to answer your question, I reckon it''s an axiom.

Like quarnin shows, it works for the complex field anyway.

Miles

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The REAL numbers are a closed, one dimensional manifold. They''re closed under the operations of addition, subtraction, multiplication and division since applying each operator to the manifold produces the same manifold.

There''s an elegant way of writing this in set theory, but it escapes me at the moment (I''m about to head out to lunch and my stomach is grumbling)!

Cheers,

Timkin

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