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Miike012
Jan4-11, 09:57 PM
1. The problem statement, all variables and given/known data

The proof has to do with factoring...... What do all the letters mean?
I added an attachment.

2. Relevant equations



3. The attempt at a solution

Apphysicist
Jan4-11, 10:06 PM
F is an arbitrary factor.
A is an arbitrary number.
B is an arbitrary number, distinct from A and F.
a and b are arbitrary numbers, used to make it easier to understand in which case for they are being used.
n and m are arbitrary numbers.

Take the first case: Given F divides A (that is, if you divide a number A by F, the result is a natural number), then prove that it divides mA (that is, prove that F divides any multiple of A).
They then use a general case that A=aF...A is a multiple of F (since F divides A, it must be a factor and thus A a multiple) to show from there.

Does that make the second case clearer?

chiro
Jan4-11, 10:09 PM
1. The problem statement, all variables and given/known data

The proof has to do with factoring...... What do all the letters mean?
I added an attachment.

2. Relevant equations



3. The attempt at a solution

The variables can be anything: ie they can have any value. The only restriction is that the variables 'a' and 'b' have to be non zero whole numbers. Besides that the variables can be any whole number you like.

Miike012
Jan4-11, 10:37 PM
Here is something else that I just came across...
the HCF of the numerator and denominator must be a factor of their sum (Thats odd isnt it? Is there some proof to this?)

The example:( 3x^3 - 13x^2 +23x -21) / 15x^3 - 38x^2 -2x +21)
Sum of num and den. = 18x^3 - 51x^2 +21x = 3x(3x-7)(2x-1)

This is where I get confused.... The book says " if there is a common divisor is is clearly 3x - 7." My problem is, the book makes it sound like it is the obvious choice..... How would you know that 3x-7 is the HCF without dividing 3x-7 then 3x then 2x-1 sepperatly into the denominator and numerator?

Apphysicist
Jan4-11, 10:48 PM
Shot in the dark:

Maybe because the coefficient of the highest degree term in each polynomial is not divisible by 2 (precluding 2x-1) and because each polynomial has a term of degree 0 so so much for 3x?

Miike012
Jan4-11, 10:52 PM
Does this "theorem" have a name? Ive never hurd of this before? Its interesting.