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?

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TopNewestEven 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 · 2 months, 1 week ago

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@Brian Charlesworth Would you help me with this? – Atomsky Jahid · 2 months, 1 week ago

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

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 · 2 months, 1 week ago

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"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 · 2 months, 1 week ago

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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 · 2 months, 1 week ago

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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 · 2 months, 1 week ago

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Thank You! – Terry Chadwick · 2 months, 1 week ago

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