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Stuck in Remainders!

Consider the following division, \[\frac{x+1}{2x+2}=\frac{\frac{1}{2}(2x+2)}{2x+2}=\frac{1}{2}\] So, dividing \(x+1\) by \(2x+2\) leaves no remainder. [Of course, \(x\) cannot be \(-1\) here.]

Now, substituting \(x=4\) in the expressions gives us two integers 5 and 10. But, \[5=0 \times 10 +5\] We're getting 5 as a remainder now!

Why are these two ways of defining remainders not congruent?

Note by Atomsky Jahid
5 months, 2 weeks ago

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Even though the same word "remainder" is used, you have to bear in mind what the definition in each of these contexts are.

Calvin Lin Staff - 5 months ago

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@Brian Charlesworth Would you help me with this?

Atomsky Jahid - 5 months, 1 week ago

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With proper fractions, (i.e., where the numerator is strictly less than the denominator), the remainder is just the numerator itself, so in your general case,

\(x + 1 = 0 \times (2x + 2) + (x + 1)\),

i.e, \(x + 1\) divided by \(2x + 1\) is \(0\) with remainder \(x + 1\). Defined this way, a proper fraction does have a remainder, which avoids the contradiction you mention.

Brian Charlesworth - 5 months, 1 week ago

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I understand your definition. But, is it the proper one? What I've found in the Wikipedia is the following.

"Given two polynomials f(x) (the dividend) and g(x) (the divisor), asserts the existence and the uniqueness of a quotient q(x) and a remainder r(x) such that \[f(x)=q(x)g(x)+r(x) ; and, r(x)=0 ,or, deg(r) < deg(g)\]"

But, \(2x+2\) and \(x+1\) both are first-degree polynomials!

Again, I solved some problems where I had to resort to the algebraic manipulation I mentioned in this note.

In an algebra problem, I had to do the following. \[R(x)=x^5+\frac{1}{2}x^3+x^2-\frac{1}{2}x+\frac{5}{2}\] \[R(x)=(x^2-\frac{1}{2})(x^3+x+1)+3\] \[R(x)=x^2(x^3+x+1)-\frac{1}{2}(x^3+x+1)+3\] Then, I had to say \(R(x)\) divided by \(x^3+x+1\) leaves a remainder of 3.

So, basically we cannot incorporate your definition, can we?

Atomsky Jahid - 5 months, 1 week ago

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@Atomsky Jahid Does the Wikipedia definition assume the \(deg(f) \gt deg(g)\)? In any event, it also depends on what field we're working with. If we are working with rationals we can indeed say that \(x + 1 = \dfrac{1}{2}(2x + 2) + 0\), which satisfies the Wikipedia definition.

In your example where you substituted \(x = 4\), this would translate to \(5 = \dfrac{1}{2} \times 10 + 0\), but this is not how we normally deal with remainders over the integers, so I think this is the source of the "incongruence" you mention in your question.

(If \(deg(f) \lt deg(g)\) then we would necessarily have \(q(x) = 0\) and \(r(x) = f(x)\), in which case the Wikipedia definition would be satisfied.)

Brian Charlesworth - 5 months, 1 week ago

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@Brian Charlesworth In our case \(deg(f) = deg(g)\). And, the definition assumes \(deg(f) \geq def(g)\). [This is also the case in Brilliant wiki about polynomial division.]

I also forgot to mention one thing about your first comment. You said, "where the numerator is strictly less than the denominator". But, notice that \(x+1\) is strictly greater than \(2x+2\) \(\forall x < -1\).

Atomsky Jahid - 5 months, 1 week ago

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Dear sir, can you please help https://brilliant.org/discussions/thread/can-this-limit-be-cancelled/?ref_id=1368476

Thank You!

Terry Chadwick - 5 months, 1 week ago

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