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## Month: February, 2015

### An alternate set-theoretic representation of n-tuples

This is just a brief note regarding the alternate set-theoretic representations of $n$-tuples.

I had read in a text on set theory that the $2$-tuple $(a,b)$ could be represented as a set in the form $\{\{a,b\},\{a\}\}$. I tried hard to think of other ways of representing such a tuple, but could not think of any then.

Today, I again went back to thinking about the issue for some reason, and came up with $\{\{a,b\},\{\{a\},b\}\}$.

What is the main issue we’re dealing with here? In a set, the order in which the elements are written does not matter. For example, $\{a,b,c\}=\{a,c,b\}=\{b,c,a\}$, etc. Hence, denoting the order in which elements appear in the tuple is tricky. The essential observation is that the reader should be able to recognize which element comes at which position in the tuple by comparing the number of parentheses it has in one element in the set as compared to the number of parentheses it has in the base element of the set (the one with the lowest number of parentheses for all elements).

For example, consider the set

$\{\{a,b,c\},\{\{a\},b,c\},\{a,\{\{b\}\},c\},\{a,b,\{\{\{c\}\}\}\}\}$.

Here the “base” element can be considered to be $\{a,b,c\}$ as it has the minimum number of parentheses for all elements. $\{\{a\},b,c\}$ has one additional set of parenthesis containing $a$. Hence, $a$ occupies the first position in the tuple. $b$ has two sets of extra parentheses containing it in $\{a,\{\{b\}\},c\}$. Hence $b$ occupies the second position in the tuple; and so on.

I can represent an $n$-tuple $(a,b,c,\dots,n)$ in the form $\{\{a,b,c,\dots,n\},\{\{a\},b,c,\dots,n\},\{a,\{\{b\}\},c,d,\dots,n\},\dots\}$.

EDIT: A representation that takes much lesser effort to write is $\{\{a, b, c, \dots, n\}, \{a\}, \{\{b\}\}, \{\{\{c\}\}\}, \dots, \}$.

### An exposition of some elementary inequalities

Short, clever proofs have played an immensely important role in the development of Mathematics. However, I can’t say I am a big fan of them. Mostly because I can never imagine myself coming up with those. Moreover, they don’t give a clear glimpse of the structure of what they prove. They just somehow manage to corroborate the conclusion of the proof with the axiomatic setup, which is not why people do Mathematics.

Today, I will try and decode some basic inequalities like $a^2+b^2\geq 2ab$ (1) and $(a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$ (2).

(1) $a^2+b^2\geq 2ab$ can also be written as $a^2+b^2\geq ab+ab$. How do we change $ab+ab$ to get $a^2+b^2$? We add $a(a-b)+b(b-a)$ to it. We’re taking the same number $(b-a)$, multiplying it by both $-a$ and $b$, and then adding the two products. One easy way to end this discussion right here is to note that $a(a-b)+b(b-a)= (a-b)^2\geq 0$. However, I’d prefer to make it even more elementary. Although I could write paragraphs on it, I’d suggest that the reader do the needful, considering all possibilities like $a<0$, $a, etc.

(2) $(ac+bd)^2=(ac+bd)(ac+bd)$. One adds $(ac+bd)[(a(a-c)+c(c-a)+b(b-d)+d(d-b)]$ to it to make it equal to $(a^2+b^2)(c^2+d^2)$. The conclusion follows directly from here, using the same argument that was given for (1).

(1) is obviously a form of the AP>GP inequality and (2) is the Cauchy-Schwartz inequality. These proofs may seem less beautiful and clever than those given in textbooks. But the hope is you’d find these proofs more helpful.

Let us take a ring $R(+,*)$. It is easy to prove that that $0*a=0$. The standard proof works this way: $0*a=(0+0)*a=0*a+0*a$. Adding $-(0*a)$ on both sides, we get $0=0*a$.

This fact could be used to prove that $(-a)*b=-(a*b)$. How? $(-a)*b+a*b=(-a+a)*b=0*b=0$. Hence, as $(-a)*b$ is the additive inverse of $a*b$, and as every element has a unique additive inverse, we have $(-a)*b=-(a*b)$.

Although proving the above is a fairly uninvolved exercise, it was never intuitive to me, unless I started to imagine $+$ as arithmetic addition and $*$ as arithmetic multiplication. This blog post is an attempt to make the aforementioned facts more easy to visualize, and perhaps intuitive.

What is $0*a$? To answer this, we need to understand what $0$ is in the context of distributivity of addition over multiplication. It is $(-a+a)$. It is $(-2a+2a)$. It is the combination of matter and anti-matter. It takes away as much as it gives. This I feel is the easiest way to think about it in the given context. I would love to elaborate on why this is the best way some other time.

Now what is $(-a)*b$? It is $(a+-2a)*b$. It is $(2a+-3a)*b$. If $(2a+-3a)*b=(a+-2a)*b$, then $a*b+(-a)*b=0$. This proves that $(-a)*b=-(a*b)$, as the additive inverse of every element is unique.

Using this fact, we can now say that

$0*a=(c+-c)*a=c*a+(-c)*a=0$.

All this because I find it easier to visualize $0$ as $(-a+a)$ for any element $a$ rather than $0+0$. And I’m sure you’d find it easier too.