Linked Questions
15 questions linked to/from cancelling before evaluation in polynomial quotients (rational functions)
59
votes
9
answers
16k
views
Has there ever been an application of dividing by $0$?
Regarding the expression $a/0$, according to Wikipedia:
In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by $0$, gives $a$ (assuming $a\not= 0$), and ...
47
votes
4
answers
18k
views
How can we prove Sylvester's determinant identity?
Sylvester's determinant identity states that if $A$ and $B$ are matrices of sizes $m\times n$ and $n\times m$, then
$$ \det(I_m+AB) = \det(I_n+BA)$$
where $I_m$ and $I_n$ denote the $m \times m$ ...
39
votes
5
answers
5k
views
Why are polynomials defined to be "formal" (vs. functions)?
Despite the fact that $\forall n, n^3 + 2n \equiv 0 \pmod 3$, I understand that $n^3 + 2n$ (considered as a polynomial with coefficients in $\mathbb Z/3\mathbb Z$) is not equal to the zero polynomial.
...
25
votes
5
answers
5k
views
How does partial fraction decomposition avoid division by zero?
This may be an incredibly stupid question, but why does partial fraction decomposition avoid division by zero? Let me give an example:
$$\frac{3x+2}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$$
Multiplying ...
4
votes
3
answers
1k
views
How to prove that if $a\equiv b \pmod{kn}$ then $a^k\equiv b^k \pmod{k^2n}$
What I have done is this:
$a\equiv b \pmod{2n}$,
$a=b+c\times2n$, for some $c$,
$a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,
$a^2-b^2=(b\times c+c^2n)\times4n$, then
$a^2\equiv b^2\pmod{2^2n}$.
...
5
votes
3
answers
5k
views
Determine the remainder when $f(x) = 3x^5 - 5x^2 + 4x + 1$ is divided by $(x-1)(x+2)$
This question arose while I was tutoring a student on the topic of the Remainder Theorem. Now, the Remainder Theorem tells us that when a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the ...
3
votes
2
answers
1k
views
gcd of $x$ and $2$ in $\mathbb Z[x]$
In $\mathbb Z[x]$, $x$ and $2$ have gcd $1$. But they cannot be expressed as the linear combination of two polynomials.
Then assuming that $1=2f(x)+xg(x)$ we are supposed to arrive at a ...
8
votes
2
answers
715
views
Definition of differentiation in context of abstract algebra.
In any regular calculus or real analysis course, we learn the definition of the derivative of a function $f(x)$ as $$f^\prime (x)=\lim\limits_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ However while ...
3
votes
1
answer
2k
views
The universal property of polynomial rings and universal identities
To verify some identity involving say two variables $x, y \in R$ for any commutative ring $R$ it suffices to verify this identity in $\mathbb{Z}[x,y]$. This is just the sort of trick that should be ...
2
votes
5
answers
325
views
How to prove that $(x-1)^2$ is a factor of $x^4 - ax^2 + (2a-4)x + (3-a)$ for $a\in\mathbb R$?
Let $a \in R$. Verify that $(x − 1)^2$ is a factor of
$$p(x) = x^4 − ax^2 + (2a − 4)x + (3 − a)$$
How can I solve this question?
3
votes
1
answer
384
views
Why is modular inverse notation so ambiguous?
Consider
$$\frac{a}{a}\pmod a,\ \ \ a\in\mathbb Z\setminus\{-1,0,1\}$$
There are two cases:
$1)$ $\frac{a}{a}$ is the notation for the real number $1$.
Then the expression is equivalent to $1$ ...
1
vote
3
answers
102
views
Very simple function question
Formally speaking, can $\frac{x}{x}$ be defined for $x=0$? Normally this division turns to be 1 but we also know that we can only divide it when the denominator is different from zero. In this case, ...
0
votes
1
answer
212
views
How does remainder theorem work even though we are technically dividing by zero?
In the polynomial remainder theorem I've noticed one thing that triggers me
$P(x) = Q(x)D(x) + R(x)$
Here we say that the divisor is $D(x) = x-a$ now to show that $P(a) = R(x)$, people simply put $x-a ...
1
vote
1
answer
166
views
Can differentiation be defined in an algebraic way?
Is it possible to define the operation D of differentiation of real functions in an abstract way, as for example by the fundamental properties of the derivative:
D(f+g) = D(f) + D(g)
D(fg) = fD(g) ...
0
votes
1
answer
100
views
Division by zero in Ireland and Rosen's "A Classical Introduction to Modern Number Theory"
In Ireland and Rosen's A Classical Introduction to Modern Number Theory (Second Edition) the proof of Chapter 6 Section 4 Proposition 6.4.2. starts with dividing $x^p-1=(x-1)\prod_{j=1}^{p-1} (x-\zeta^...