Negative numbers from pairs
Tonight's idea
Pretend the only numbers are 0, 1, 2, 3 and so on. You can always add, but you cannot always take away: \(3 - 5\) has no answer yet.
Write a pair \((a, b)\) and read it as "\(a\) steps forward, then \(b\) steps back". Then \((5, 3)\) and \((7, 5)\) both leave you 2 steps ahead, so they should count as the same. You want to say: the same when \(a - b = c - d\). But taking away is what we are missing. So add \(b\) and \(d\) to both sides.
Nothing here uses subtraction, so every pair is allowed, even \((3, 5)\). It matches no ordinary number. It is a new one: 2 steps behind the start.
Pairs used like this are called formal differences. In the 1950s Alexander Grothendieck used the same idea, with a slightly adjusted rule, on things much stranger than numbers, and that is where a subject called K-theory begins.
Watch one: three names for 2, and a new number
Same or not? Test \((5, 3)\) against \((7, 5)\): outside numbers \(5 + 5 = 10\), inside numbers \(3 + 7 = 10\). Equal, so they are the same. Test \((5, 3)\) against \((2, 0)\): \(5 + 0 = 5\) and \(3 + 2 = 5\). The same again. So \((5, 3)\), \((7, 5)\) and \((2, 0)\) are three names for one number, the ordinary number 2. In general the ordinary number \(n\) is the pair \((n, 0)\).
Order matters. Test \((5, 3)\) against \((3, 5)\): \(5 + 5 = 10\) but \(3 + 3 = 6\). Not the same.
The sum with no answer. The number 3 is \((3, 0)\). Taking away 5 means adding 5 steps back, which is \((0, 5)\). Add part by part, firsts with firsts and seconds with seconds:
\[(3, 0) + (0, 5) = (3 + 0,\; 0 + 5) = (3, 5).\]Is there a simpler name? Test \((3, 5)\) against \((0, 2)\): \(3 + 2 = 5\) and \(5 + 0 = 5\). The same. So \(3 - 5\) now has an answer, the pair \((0, 2)\): 2 steps behind the start. Today we write it \(-2\).
Your turn
- Which of these pairs are the same as \((3, 8)\)? Use the rule \(a + d = b + c\) and show both sums each time. \((0, 5)\), \((8, 3)\), \((5, 10)\), \((13, 18)\), \((4, 8)\).
- Every pair has a simplest name, with a 0 in it. Find the simplest name for \((9, 4)\), \((4, 9)\), \((10, 25)\) and \((6, 6)\). Which of the four are ordinary numbers?
- Add \((2, 5) + (4, 1)\) and give the simplest name of the answer. Then add \((6, 6)\) to \((9, 4)\) and check, with the rule, that the result is the same as \((9, 4)\).
- Write two different pairs that both mean "4 steps behind the start". Then find a pair that undoes \((7, 2)\): added to \((7, 2)\), it gives a pair that is the same as \((0, 0)\). What is the quick way to undo any pair \((a, b)\)?
- A fairness test. \((5, 3)\) and \((2, 0)\) are two names for one number. Add \((1, 4)\) to each of them. Are the two answers the same as each other? Why would the whole idea fall apart if they were not?
Stretch: the same trick with times
Go back to the numbers 0, 1, 2, 3 and so on. You can multiply any two, but you cannot always divide. So make new pairs \((a, b)\), with \(b\) not 0, and call \((a, b)\) and \((c, d)\) the same when \(a \times d = b \times c\). Check that \((2, 3)\) and \((4, 6)\) are the same and that \((1, 2)\) and \((2, 1)\) are not. What do you usually call these pairs? Work out \((1, 2) \times (1, 2)\) and \((1, 2) + (1, 2)\), both part by part. Which answer is sensible? Last, why must \((0, 0)\) be banned? Test it against \((1, 2)\) and against \((1, 3)\).
Answers
- \((0, 5)\): \(3 + 5 = 8\) and \(8 + 0 = 8\). The same. \((8, 3)\): \(3 + 3 = 6\) and \(8 + 8 = 16\). Not the same. It is \((3, 8)\) backwards, 5 steps ahead instead of 5 behind. \((5, 10)\): \(3 + 10 = 13\) and \(8 + 5 = 13\). The same. \((13, 18)\): \(3 + 18 = 21\) and \(8 + 13 = 21\). The same. \((4, 8)\): \(3 + 8 = 11\) and \(8 + 4 = 12\). Not the same. So three of the five match: \((0, 5)\), \((5, 10)\) and \((13, 18)\).
- \((9, 4)\) is \((5, 0)\), because \(9 + 0 = 4 + 5\). \((4, 9)\) is \((0, 5)\), because \(4 + 5 = 9 + 0\). \((10, 25)\) is \((0, 15)\), because \(10 + 15 = 25 + 0\). \((6, 6)\) is \((0, 0)\), because \(6 + 0 = 6 + 0\). The ordinary numbers are the ones whose 0 comes second: \((5, 0)\) is 5 and \((0, 0)\) is 0. The other two are new: 5 behind and 15 behind, which we now write \(-5\) and \(-15\). To find a simplest name, take the smaller number of the pair off both parts.
- \((2, 5) + (4, 1) = (6, 6)\), whose simplest name is \((0, 0)\). That is zero: 3 behind plus 3 ahead. Then \((6, 6) + (9, 4) = (15, 10)\). Test against \((9, 4)\): \(15 + 4 = 19\) and \(10 + 9 = 19\). The same. Adding \((6, 6)\) changes nothing, just as adding 0 should.
- Any pair whose second number is 4 more than its first will do: \((0, 4)\), \((1, 5)\), \((6, 10)\) and so on. To undo \((7, 2)\) use \((2, 7)\): \((7, 2) + (2, 7) = (9, 9)\), and \(9 + 0 = 9 + 0\), so that is the same as \((0, 0)\). Any other name for \((2, 7)\), such as \((0, 5)\), works too. The quick way: swap the pair. \((a, b) + (b, a) = (a + b,\; a + b)\), which is always the same as \((0, 0)\). Every pair can be undone, and that is exactly what the ordinary numbers were missing.
- \((5, 3) + (1, 4) = (6, 7)\) and \((2, 0) + (1, 4) = (3, 4)\). Test: \(6 + 4 = 10\) and \(7 + 3 = 10\). The same, both 1 behind. If they had come out different, the answer to a sum would depend on which name you happened to pick for a number, and then the pairs would not behave like numbers at all. It works every time: adding \((1, 4)\) to both names puts an extra \(1 + 4\) on each side of the test, so equal sums stay equal.
Stretch. \(2 \times 6 = 12\) and \(3 \times 4 = 12\), so \((2, 3)\) and \((4, 6)\) are the same. \(1 \times 1 = 1\) but \(2 \times 2 = 4\), so \((1, 2)\) and \((2, 1)\) are not. These pairs are fractions: \((a, b)\) is \(\tfrac{a}{b}\), and the rule is the cross-multiplying test for equal fractions. Multiplying part by part gives \((1, 2) \times (1, 2) = (1, 4)\), and a half of a half is a quarter. Sensible. Adding part by part gives \((2, 4)\), which is the same as \((1, 2)\), so a half plus a half would be a half. Not sensible. Part by part belongs to the operation you started with, which is times here. Fractions need a different rule for adding. Finally \((0, 0)\): \(0 \times 2 = 0 \times 1\) and \(0 \times 3 = 0 \times 1\), so it would be the same as both \((1, 2)\) and \((1, 3)\). But those two are different, since \(1 \times 3\) is not \(2 \times 1\). Two different numbers would both equal a third, and "the same" would stop making sense. That is one reason a fraction never has 0 underneath.