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In 'Mathematics Form and Function' ch.2, section 4, 'Integers' by Saunders Mac Lane (p.50 in the 96 edition) I came across the following definitions of sum and product for ordered pairs:

(m, n) + (m', n') = (m + m', n + n')
(m, n) (m', n') = (mm' + nn', mn' + m')

My understanding is that each m or n are related to m' and n' in such a way that, for example, if m' is m + 1, then n' must be n + 1; the relationship must be consistent between both the original and subsequent numbers.

Given that, if I plug in some real numbers such as m = 2 and n = 3 (and in this case the number with the prime symbol is equivalent to adding 1), I would get this:

(2 + 3, 3 + 4) = (5, 7) # still obtain an ordered pair, so far so good
((2 * 3) + (3 * 4), (2 * 4) + 3) = (18, 11) # not an ordered pair!

My understanding is the result of the product should be an ordered pair with the 2nd element greater than the 1st, but that's the not case because 18 > 11. Is this an error, typo, or am I fundamentally misunderstanding the maths here?

Thanks for any advice you can give.

zhanmusi
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2 Answers2

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The word ordered in ordered pair $(a,b)$ does not mean $a<b$

It simply means that $a$ is the first element and $b$ is the second one.

The difference between the set and ordered pair is that $\{a,b\}=\{b,a\}$ but $(a ,b)\ne (b,a)$

Mohammad Riazi-Kermani
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The notion 'ordered pair' simply refers to a sequence of any elements of length 2, independently from any ordering among the elements.

So, $(18,\,11)$ is a perfect ordered pair.

'Ordered' here only means that the order of the two elements is important, so that $(18,\,11)\ne(11,\,18)$.

Also, in the definition, $m, n, m', n'$ are all arbitrary, there is no assumption on any relations among them.

Berci
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  • Thanks for the explanation. Are you able to explain the definition of the product in more detail? I'm afraid I don't understand why it is defined as it is (whereas the definition of the sum seems intuitively correct to me). – zhanmusi Aug 04 '19 at 20:28
  • It is not a generic definition of product for ordered pairs, but a specific one, probably only for the given exercise.. – Berci Aug 04 '19 at 20:49
  • By the way, isn't the rule for product rather $(m,n)\cdot (m', n') =(mm'-nn',, mn'+m'n)$? That's how multiplication works for complex numbers.. – Berci Aug 04 '19 at 21:37
  • Your last point I think is the key @Berci. The rule for product as you list it there is what I expected (although not based on experience, but thinking logically I expected n to also appear in the 2nd part of the ordered pair). I think the ultimate answer is to this part of my question "Is this an error, typo . . .?" I think this is a typo / error in the book as I have copied it exactly. If you incorporate the correct rule for product into your answer then I can accept it. – zhanmusi Aug 04 '19 at 21:47
  • Well, one can investigate further operations on the set of pairs. For example, why not just take the natural definition $(m,n) (m', n') =(mm', nn') $? Or, there are complex-like numbers, with strange-looking multiplication.. (However, that $n$ might indeed be missing from the formula.) – Berci Aug 04 '19 at 21:59
  • Okay, I think I may have 'jumped the gun' yet again. The natural definition you give above seems most reasonable to me while the book claims to be providing one of the 'definitions' of product, & that's assuming it is correctly printed or not. I must read more to understand this as I have some assumptions which have been revealed by you & @Mohammad to be incorrect. Thanks to you both for taking the time to answer. I will wait a little while before accepting an answer but I appreciate both your efforts. – zhanmusi Aug 05 '19 at 00:35