math algebric's equality expressions

Discussion in 'Math' started by Ryan$, Jan 5, 2019.

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  1. Ryan$

    Thread Starter Member

    Dec 14, 2018
    Hi guys, I have recently faced something in programming which I think it's deeply relevant to math;
    if I have a statement in a specific pattern ... it maybe modified as totally another statement which is equal to my original statement .. for example i<x+5 is the same to say i-x<5 .. my question, is there any website that can deal with subject of equality in algebric?! thanks!!
  2. WBahn


    Mar 31, 2012
    Uh... how about any 7th grade algebra text book?
  3. Ryan$

    Thread Starter Member

    Dec 14, 2018
    May please have a look on my thread over general discussion aka " Time complexity cumulative sum?"
    there i have the same probl with multiply and dividing
  4. danadak

    Well-Known Member

    Mar 10, 2018
    I am not a math expert by any measure. But I always found fascinating
    the algebraic re-arrangement of expressions which led to broader insights
    into physical behavior and the properties of systems, objects, etc..

    Same for ratio arrangement.

    These expressions, i<x+5, i-x<5, the first statement tells me that the upper boundary
    of i is x + 5, whereas the second tells me the boundary of the difference of i-x is 5.
    Different insights out of the same fundamental statement. Good stuff.

    Regards, Dana.
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  5. WBahn


    Mar 31, 2012
    Please include a link to the thread (and preferably post) you are wanting someone to take a look at. You have shotgunned the forum with so many threads that few people, myself included, are going to sift through them hoping to stumble across the one you mean.

    If you are struggling with basic algebra concepts, then you are going to continue to struggle and have a hard time understanding most things that are based on them until and unless you take a big step back and address your difficulties with basic algebra concepts. The longer that you choose not to do so is just that much more time you are going to spend digging yourself a hole that will be increasingly difficult to every get yourself out of.
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  6. bogosort

    Active Member

    Sep 24, 2011
    If you want to understand the math behind equality (and inequality), research binary relations. In short, a binary relation is like a function with no restrictions. Recall that a function f:A → B is a set of ordered pairs (a, b), such that a ∈ A and b ∈ B, with the property that every element in A is associated to exactly one element in B. That property is a restriction on the kinds of things we can call functions. In contrast, a binary relation R:A → B is a set of ordered pairs (a, b) with no restrictions. So, binary relations are more general than functions. For example, taking the square root of a number is not a function -- the number 4 is associated with two square roots, +2 and -2 -- but it is a relation, as both (4, 2) and (4, -2) are allowed in a relation.

    Equality is a special kind of binary relation called an equivalence relation. Let R be a relation on some set X, with x, y, z ∈ X; then, R is an equivalence relation if and only if it has the following properties:
    • reflexivity: (∀x)[(x, x) ∈ R]
    • symmetry: (∀x∀y)[(x, y) ∈ R → (y, x) ∈ R]
    • transitivity: (∀x∀y∀z)[(x, y) ∈ R ∧ (y, z) ∈ R → (x, z) ∈ R]
    Identifying R with the symbol '=', these correspond, respectively, to
    • x = x
    • if x = y, then y = x
    • if x = y and y = x, then x = z
    These properties formalize the notion of what we intuitively think of as "sameness" between elements. In the case of sets of elements, we can formalize what it means for two sets A and B to be equal:

    A = B ↔ (∀x)([x ∈ A → x ∈ B] ∧ [x ∈ B → x ∈ A])​

    In other words, A equals B if and only if every element in A is an element in B and every element in B is an element in A. I won't bother proving it here, but it can be shown that this definition of set equality is an equivalence relation, and so the three properties above apply.

    In your example, let R1:ℤ → ℤ be the set of order pairs (x, y) such that y < x + 5. R1 is thus a binary relation on the integers. Likewise, let R2:ℤ → ℤ be the set (x, y) such that y - x < 5. R2 is also a relation on the integers.

    R1 has elements such as (5, 8) and (5, 9), but not (5, 10), as 10 is not less than 5 + 5. Likewise, R2 has elements such as (5, 8) and (5, 9), but not (5, 10), as 10 - 5 is not less than 5. Indeed, every element in R1 is an element in R2, and every element in R2 is an element in R1, therefore R1 = R2. We knew this beforehand algebraically, by manipulating the symbols, but the underlying mathematical reason for their equality is that R1 and R2 describe precisely the same set.
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014

    Perhaps it might help to look into Lambda Calculus. Not entirely sure what you are asking here though.
  8. drc_567

    AAC Fanatic!

    Dec 29, 2008
    ... Sometimes a graphical interpretation is possible. For example, the expression X+5 can be expressed as X+5=0, which, solving for X, gives X=-5. When plotted using Cartesian Co-ordinates, that term is plotted as a vertical line at the position X=-5. Then, the original inequality, i<X+5, is the region on the graph to the left of the line at X=-5.
    Last edited: Jan 13, 2019
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