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Given two sets $A$ and $B$ of cardinality $a$ and $b$, respectively, the cardinality of the cartesian product $A\times B$ is called the product of $a$ and $b$, and is denoted by $a\cdot b$. Update When I wrote this answer I didn't have infinite sets in mind. I just wanted to convey a mental picture of multiplication that does not involve repeated ...


$a\cdot b$ is the the value of $f_a(b)$, where $f_a$ is the unique endomorphism of $\mathbb N$ (under addition) satisfying $f_a(1)=a$.


As k.stm says in the comments, usually a more general thing is true: these categories $C$ are equipped with forgetful / underlying set functors $U : C \to \text{Set}$ which tend to have a left adjoint, the "free" functor $F : \text{Set} \to C$. Whenever this is true, it follows that $U$ preserves all limits, not just products. Sometimes, but more rarely, ...


Here is an answer that requires an $x$ and $y$ axis. Let us say that we would like to multiply $A$ and $B$. Then we locate the point, $(1,A)$ and make the line determined by $(0,0)$ and $(1,A)$. Then locate the point $(B,0)$, and draw the vertical line that goes through this point. Then find the intersection of the vertical line just formed and the line ...


We can define $\mathbb{N}$ as the initial semiring. In this approach, not only do we not have to define multiplication as repeated addition, but, in fact, we do not have to define multiplication at all. :)


If you want a definition of the usual operation of multiplication that does not reduce it to repeated addition, you can always define it axiomatically the same way that we were taught it: If $a, b$ are between 0 and 9, $a\cdot b$ is given by the multiplication table. Otherwise, the operation is defined recursively by specifying the rules of digit-by-digit ...


Combining my comments into an answer: First define $a^2=\sum_{i=1}^a (2i-1)$. Then define $a\cdot b={(a+b)^2-a^2-b^2\over 2}$. (You could of course argue that this is equivalent to repeated addition, but the same would be true of any valid definition.)


The binary operation $\cdot$ is multiplicative if There exists an $x$ such that, for every $y$, $x\cdot y=y$. That is, there is a (left?) identity element. There exists an additive binary operation $+$ such that for every $x$, $y$, and $z$, $x\cdot(y+z)=x\cdot y+x\cdot z$. That is, $\cdot$ is (left?) distributive over $+$. This, of course, requires a ...


The question can be rephrased to: why does the forgetful functor so often preserve limits? (Because in all of the cases you mention equalisers are preserved too.) This property of the forgetful functor is forced by abstract nonsense (representable functors are continuous) and by the fact that forgetful functors very often are representable. As for why the ...


I like the definition of $ab$ as the area of a rectangle with side lengths $a$ and $b$. Then it's clear that multiplication is commutative, for example. Incidentally, this can sometimes be useful in communicating with non-mathematicians. For example, you can say that a pile of stones has a prime number of stones in it if there's no way to arrange them into ...


Define set $S_n$ of multiples of $n$ as intersection of all sets $S\subseteq\Bbb N$ such that $0\in S,\forall m\in S:m+n\in S$. Define least common multiple of $a,b$ as the least positive element of $S_n\cap S_m$. Define $n^2$ as a number $n$ less than lcm of $n,n+1$. Define $a\cdot b$ as half of the number $(a+b)^2-a^2-b^2$.


Hint: There is an identity of the form $\cos 3 \alpha = \cos \alpha \cdot (2 \cos 2 \alpha - 1)$. Multiply this as $\alpha$ runs through $2^k \pi/13$ with $0 \leq k < 6$. Using standard properties of $\cos \alpha$, you will be able to cancel out the redundant terms.


You have $\cos 3\alpha=\cos \alpha(2\cos 2\alpha-1)$ You have also $\cos \frac{32\pi}{13}=\cos \frac{6\pi}{13}$ $\cos \frac{64\pi}{13}=\cos \frac{12\pi}{13}$ $\cos \frac{16\pi}{13}=\cos \frac{10\pi}{13}$ Thus at the end you have $\prod_{i=1}^6 \left(2\cos\left(\frac{2^{i}\pi}{13}\right)-1\right)=\prod_{i=1}^6 \dfrac{\cos \frac{3i\pi}{13}}{\cos ...


I'm not sure if it's exactly what you're looking for, but just about any intro to combinatorics (enumerative combinatorics) would contain a lot of these kinds of things. One book that comes to mind is "Discrete Mathematics and its Applications" by Rosen. It is a text for intro proof writing (so it'll contain induction and the like) and will certainly see a ...


$$a\times b = \begin{cases} \frac{a}{2} \times \left(2\times b\right) &\text{ if } a \text{ is even}\\ b +\frac{a-1}{2}\times\left(2\times b\right)& \text{ if } a \text{ is odd} \end{cases}$$


Call your sum $S$. Note first that if $n=1$, $S=2^k$, which is not prime. If $n =2^k-1$, then $S =2$ is prime (note that this holds iff). Assume now that $n$ is a proper factor of $2^k-1$ ($2^k-1$ is no Mersenne Prime n, in which case $S$ could only be an integer for $n=2^k-1)$. Now, $2^k-1$ has no factor $2$, so it is a product of odd primes. If $n$ is a ...


Here is a more "1st principles" pf. I use a hint in Marsden's book. 1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas. Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy $(1-z)^n=1 \leftrightarrow (1-z) \in \{\cos \frac{2 \pi k}{n}+i \sin \frac{2\pi k}{n}:k=0,...,n-1 \}$ (the set of nth roots ...

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