My Math Forum Properties of non-integer combinations

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 November 26th, 2017, 08:57 PM #1 Newbie   Joined: Nov 2017 From: India Posts: 3 Thanks: 0 Properties of non-integer combinations A linear combination, ax+by+...=z, where a,b,...,z∈Z and x,y,...∈Z. It is bound to yield an integer by the closure property of integers under the addition operation. This fact is used in computing g.c.d. among others. I want to know about properties of non-integer combination, i.e. given a,b,...∈Z; but the multipliers x,y,.. not all ∈ Z. I hope that they must be enjoying similar properties, as they are comprised of rationals, and the rationals are closed under addition too. If so, then how the property (of closure of rationals under addition) can be used where the linear combinations do not hold. A familiar case is in g.c.d. computation, where the invariant property (g.c.d. is same at each step) is a product of two linear equations being followed at each step that lead to common divisors of two pairs: (i) remainder (r), divisor(a), & (ii) dividend (d), and divisor(a). The Euclid algorithm reduces the quantities of dividend(d), divisor(a) at each step, while keeping the invariant property being followed. In (i) & (ii), d & r are the linear combinations respectively. Quotient (q) and divisor (a) are not linear combinations as when taken on l.h.s. lead to r.h.s. side expressions of d−r/a & d−r/q respectively. Definitely the expression given by d-r/q or d-r/a is a rational expression, and rationals are closed w.r.t. to the addition operation. I want to know, as curiosity, what properties are enjoyed by the two quantities that are not linear combination of integers.
 November 28th, 2017, 04:09 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 I am not clear what you are asking. You start by saying "multipliers x,y,.. not all ∈ Z". From just that they could be irrational. But then you say "as they are comprised of rationals". That sounds like you are thinking that is they are not integers, they must be rational, which is, of course, not true. Do you mean that you are requiring these numbers to be rational? If so then for any sum, ax+by+...=z, you can multiply by the least common denominator of the rational number to get back to the integer case. Any thing that is true of such a sum with integer multipliers is true with rational number multipliers. Thanks from greg1313 and v8archie
 November 28th, 2017, 06:26 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,446 Thanks: 2499 Math Focus: Mainly analysis and algebra The GCD is not a function with an equivalent in the rationals either. Thanks from greg1313
November 29th, 2017, 07:22 PM   #4
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 Originally Posted by Country Boy I am not clear what you are asking. You start by saying "multipliers x,y,.. not all ∈ Z". From just that they could be irrational. But then you say "as they are comprised of rationals". That sounds like you are thinking that is they are not integers, they must be rational, which is, of course, not true. Do you mean that you are requiring these numbers to be rational? If so then for any sum, ax+by+...=z, you can multiply by the least common denominator of the rational number to get back to the integer case. Any thing that is true of such a sum with integer multipliers is true with rational number multipliers.
I meant rationals, i.e. a super-set of integers. Irrationals never crossed my mind, so I am sorry for confusion. I hope you are correct that rationals can be converted to integers. But, the moot question still remains that the rational expression is not a linear combination. My question was specifically in reference to the fraction (d - r)/q or (d -r)/a being a rational and not an integer. So, it is not enjoying any properties that are enjoyed by the linear combinations.

Last edited by ajiten; November 29th, 2017 at 07:27 PM.

November 29th, 2017, 07:30 PM   #5
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 Originally Posted by v8archie The GCD is not a function with an equivalent in the rationals either.
My question was specifically in reference to the fraction (d - r)/q or (d -r)/a being a rational and not an integer. I meant that the properties of linear combinations are not enjoyed by a rational. So, a and q can not be expressed as a linear combination of the other terms.

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