Mathematics Asked on January 8, 2021
I have been reading Tao’s Analysis I, Chapter 4 which he builds up the definition of integers and in particular, the definition of subtraction using a placeholder “ $ —$” such that $a —b$ implicitly implies $a-b$.
He goes on to define several definitions of the subtraction placeholder and its properties using the previous chapters before coming to the conclusion that the placeholder and the minus sign of subtraction (as we know it) are equivalent to one another.
I am a little sceptical with this way of looking at subtraction from Tao’s foundational standpoint. It seems that the logic is circular in the fact that subtraction is already assumed before defining it, thus rendering all the sidelines of placeholders unnecessary. One could merely take subtraction as it is common known and show its properties. Why did Tao them require the placeholder for subtraction in the first place?
I also do know that he at Pg. 74 mentioned that Tao had admitted integers, defined as what you can get by subtracting two natural numbers to be circular. This I guess is pointing towards the same line of argument as I am making?
Also, proposition 2.2.6, defining the cancellation property (i.e given $a+b = a+c$ thereby $b=c$ ) already a definition of subtraction by itself? Why did he still take it as a “virtual subtraction” when by itself it could be well defined as a core of all subtraction cases?
What am I missing out here?
Is there a better way to look at the idea of subtraction from a less circular point of view?
As stressed by the author, the place-holder is not subtraction (still to be defined): it is written as: $(a−−b)$.
Usually (see 2nd line page 75) we write $(a,b)$ and we mean exactly that the new defined entity is a pair of naturals.
We define rules for "manipulating" pair and we check that they give us the expected properties of integers.
The first step is to verify equality: $(a,b)=(c,d)$ is defined as $a+d=c+b$. As you can see, the RHS of the definition does not use the "$-$" symbol at all.
Thus, no circularity.
Proposition 2.2.6 (Cancellation law) is about natural numbers:
Let $a, b, c$ be natural numbers such that $a + b = a + c$. Then we have $b = c$.
It is not a "full" subtraction: from it we cannot retrieve: $(a+b)-b=a$, simply because subtraction is not defined in $mathbb N$.
The next steps are:
(i) defining sum for the new entities: $(a,b)+(c,d)=((a+c),(b+d))$ (and verifying that the definition is consistent).
(ii) defining the "inverse" of new entities: $-(a,b)=(b,a)$.
(iii) "identify" the natural $n$ with the new entity $(n,0)$, verifying (see page 77) that $(n,0)=(m,0)$ if and only if $n = m$ (using property above: $n+0=m+0$).
(iv) prove that $(n,n)=(0,0)$ for every $n$ (because: $n+0=n+0$).
Now we can start checking that the new entities have the expected properties of integers: is is true that $(a,b)+ [-(a,b)] = (0,0)$ ?
Let check it: $(a,b)+ [-(a,b)] = (a,b) + (b,a)$, by (ii), $=((a+b),(b+a))$, by (i), $=((a+b),(a+b))$, by property of naturals, $=(0,0)$, by (iv).
Answered by Mauro ALLEGRANZA on January 8, 2021
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