Multiples that swallow everything
Tonight's idea
Tonight "number" means a whole number: positive, negative or zero. The multiples of 3 are ..., \(-6\), \(-3\), 0, 3, 6, ... Add two of them and you get another one. Multiply one by any number, even one that is not a multiple of 3, and you still get a multiple of 3. The set swallows multiplication.
A set of whole numbers is an ideal if it contains 0, adding two members always gives a member, and multiplying a member by any whole number always gives a member.
The multiples of any one number form an ideal. They are what a clock ignores: on a 3-hour clock, numbers that differ by a multiple of 3 count as the same.
Every multiple of 4 is a multiple of 2, so the multiples of 2 are the bigger set. Smaller number, bigger set. The multiples of 64, 32, 16, 8, 4, 2 and 1 make a growing chain of seven sets.
Starting from the multiples of a positive number, a growing chain of sets of multiples must stop. A shrinking chain need not.
Watch one: why the 3-hour clock works
The multiples of 3 are an ideal. Zero is \(3 \times 0\). Two multiples of 3 look like \(3a\) and \(3b\), and \(3a + 3b = 3 \times (a + b)\). For any number \(c\), \(3a \times c = 3 \times (a \times c)\). Try it: \(12 + (-21) = -9\) and \(12 \times 7 = 84 = 3 \times 28\).
The odd numbers are not. Zero is missing. Also \(3 + 5 = 8\) and \(3 \times 2 = 6\), both even. One failure is enough.
The clock. On a 3-hour clock, 7 and 10 count as the same, because they differ by 3. So do 5 and 11, which differ by 6. Is \(7 \times 5\) the same as \(10 \times 11\)? Write \(10 = 7 + 3\) and \(11 = 5 + 6\):
\[10 \times 11 = (7 + 3) \times (5 + 6) = 7 \times 5 + 7 \times 6 + 3 \times 5 + 3 \times 6.\]The last three pieces are 42, 15 and 18. Each is a multiple of 3 times some number, so each is a multiple of 3: that is swallowing. Their sum, 75, is a multiple of 3: that is adding. So \(10 \times 11 = 110\) and \(7 \times 5 = 35\) differ by \(75 = 3 \times 25\). Both leave remainder 2 when you divide by 3. Swap in any numbers that count as the same and the product still counts as the same.
Your turn
- Which of these sets are ideals? For each one that is not, give an example that breaks a rule. (a) The even numbers. (b) The multiples of 4. (c) The squares 0, 1, 4, 9, 16, ... (d) The odd numbers. (e) The numbers whose last digit is 0. (f) The set that contains only 0.
- Look at all the numbers \(4a + 6b\), where \(a\) and \(b\) can be any whole numbers, negative ones included. (a) Work out \(4a + 6b\) for \((a, b) = (1, 1)\), \((2, -1)\), \((-1, 1)\), \((3, -2)\) and \((-2, 1)\). (b) Why can \(4a + 6b\) never be odd? (c) Since \(2 = 4 \times (-1) + 6 \times 1\), you can make any even number. Make 14 and \(-8\). (d) Now ban negatives: \(a\) and \(b\) must be 0 or more. Which even number from 0 to 20 can you no longer make?
- Here \(m\) and \(n\) are positive. (a) The multiples of 180 contain every multiple of 360. The multiples of 7 do not. Explain both. (b) Finish the rule: the multiples of \(m\) contain the multiples of \(n\) exactly when ... (c) Build the longest growing chain of sets of multiples you can, starting with the multiples of 360. Each set must contain the one before and be bigger. How many sets are in it? (d) Use \(360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5\) to say why no chain is longer.
- Now go the other way: the multiples of 2, then of 4, of 8, of 16, and so on, doubling each time. (a) Why does each set contain the next one? (b) Name a number that is in the multiples of 8 but not in the multiples of 16. Why does this shrinking chain never stop? (c) Which numbers are in every set of the chain?
Stretch: two numbers, one ideal
Let \(a\) and \(b\) be any whole numbers, negative ones included. (a) Which numbers can be written as \(6a + 10b\)? (b) Show that the set of all numbers \(6a + 10b\) is an ideal. (c) Which numbers do you get from \(6a + 15b\)? From \(4a + 9b\)? Guess the rule for \(ma + nb\) when \(m\) and \(n\) are positive.
Answers
- (a) Yes: the multiples of 2. (b) Yes. (c) No: \(1 + 4 = 5\) is not a square, and \(4 \times 2 = 8\) is not either. (d) No: 0 is missing, \(1 + 3 = 4\) is even, and \(3 \times 2 = 6\) is even. (e) Yes: these are 0, 10, \(-10\), 20, \(-20\), ..., the multiples of 10. (f) Yes: \(0 + 0 = 0\) and \(0 \times c = 0\). It is the multiples of 0, and it sits inside every other ideal.
- (a) 10, 2, 2, 0 and \(-2\). (b) \(4a + 6b = 2 \times (2a + 3b)\), which is even. (c) Multiply \(2 = 4 \times (-1) + 6 \times 1\) by 7: \(14 = 4 \times (-7) + 6 \times 7\). Multiply it by \(-4\): \(-8 = 4 \times 4 + 6 \times (-4)\). Other answers work too, such as \(14 = 4 \times 2 + 6 \times 1\). The same trick makes any even number, so the numbers \(4a + 6b\) are exactly the multiples of 2. (d) Only 2. You can make 0, 4, 6, \(8 = 4 + 4\), \(10 = 4 + 6\), and every even number after that by adding more 4s. But 2 is not 0, and one 4 or one 6 is already too big.
- (a) A multiple of 360 is \(360k = 180 \times 2k\), so it is a multiple of 180. But 360 itself is not a multiple of 7, since \(360 = 7 \times 51 + 3\). (b) ... exactly when \(m\) divides \(n\), which means \(n\) is a multiple of \(m\). (c) Seven sets, for example the multiples of 360, 180, 90, 45, 15, 5 and 1. Your chain may differ: there are 60 chains of seven. (d) By the rule, each new number divides the one before and is smaller, so each step removes at least one prime factor. 360 has six prime factors, counting repeats. So there are at most six steps, which is seven sets.
- (a) A multiple of 4 is \(4k = 2 \times 2k\), a multiple of 2. In the same way a multiple of 16 is \(16k = 8 \times 2k\), and so on. (b) 8 itself (or 24, or \(-8\)). Every number in the list does this: 32 is in the multiples of 32 but not in the multiples of 64. So each set is truly smaller than the one before, and the doubling never runs out. (c) Only 0. A multiple of 128 other than 0 is at least 128 away from 0, so 100 is not one. In the same way, every number except 0 has dropped out by the time the doubling passes it.
Stretch. (a) Exactly the even numbers. \(6a + 10b = 2 \times (3a + 5b)\) is even, and \(2 = 6 \times 2 + 10 \times (-1)\), so \(2k = 6 \times 2k + 10 \times (-k)\). (b) Zero is \(6 \times 0 + 10 \times 0\). Adding: \((6a + 10b) + (6c + 10d) = 6(a + c) + 10(b + d)\). Swallowing: \((6a + 10b) \times c = 6(ac) + 10(bc)\). (c) \(6a + 15b\) gives the multiples of 3, using \(3 = 6 \times (-2) + 15 \times 1\). \(4a + 9b\) gives every whole number, using \(1 = 4 \times (-2) + 9 \times 1\). The rule: \(ma + nb\) gives exactly the multiples of the biggest number that divides both \(m\) and \(n\). That is true for all positive \(m\) and \(n\), though you have only tested it here. In fact every ideal of whole numbers is the multiples of one number, which is why chains of ideals behave like chains of divisors.