Cahit Arf Matematical Days Second Day May 8, 2002 N stands for the set of natural numbers like 0, 1, 2, 3 etc. Z stands for the set of integers like 2, 1, 0, 1, 2 etc. Q stands for the set of rational numbers. R stands for the set of real numbers. We define: Q[2] = {a + b2 : a, b Q}. Z[2] = {a + b2 : a, b Z}. A subset A of R is said to be closed under substraction if for all elements a and b of A, the difference a b is also in A. For example the set of integers is closed under substraction, but not the set of positive integers. A subset A of R is said to be closed under addition if for all elements a and b of A, their sum a + b is also in A. For example the set of positive integers is closed under addition, but not the set of odd integers. A subset A of R is said to be closed under multiplication if for all elements a and b of A, their product ab is also in A. A subset A of R is said to be closed under squaring if for all elements a of A, its square a2 is also in A. A subset A of R is said to be closed under inversion if for all elements a 0 of A, its multiplicative inverse a1 is also in A. You may need the following theorem that you are allowed to use: Theorem: If n N, ao, a1, ..., an Z and ao + a1 + a22 + ... + ann = 0 then ao = a1 = a2 = ... = an = 0. You have the right to use the answers to previous questions (even if you were not able to prove them) to answer next questions. 1. Show that a subset of R which is closed under substraction is also closed under addition. Answer: x + y = x ((x x) y). 2. Show that a subset of R which is closed under substraction and squaring and that contains one half of each of its elements is closed under multiplication. Answer: xy ( x y )2 x 2 y 2 . 2 3. Show that a subset of R which is closed under substraction and squaring and that contains 1/2 is closed under multiplication. Answer: Since 1/2 is in the set, so is 1/4. Now note the following: x / 2 ( x 1 / 4) 2 x 2 (1 / 4) 2 , and conclude by using part 2. 1 4. Find a subset of R which is closed under substraction and squaring but which is not closed under multiplication. Answer: {a1 + a22 + 2a33 + a44 + ... + ann : n N, ai Z } 5. Show that a subset of R which is closed under substraction and inverting and that contains 1/2 is closed under multiplication. Answer: First note that 1 is in the subset. Next note the following: (( y (1 y 1 ) 1 ) 1 y 1 ) 1 y 2 and apply part 3. 6. Show that a subset of Q which is closed under substraction and squaring is closed under multiplication. Answer: Let A be such a set. Let u, v A. Write u = a/b ve v = c/d where a, b, c, d Z. Let e be the greatest common divisor of ad and bc. Then u = a/b = (e/bd)(ad/e) (e/bd)Z. and v = c/d = (e/bd)(bc/e) (e/bd)Z. 2 2 2 Thus uv (e /b d )Z. Thus, in order to prove that uv is in A, we need to show that e2/b2d2 is in A, and for that we need to show that e/bd is in A. Since e = gcd(ad, bc) there are integers x and y such that adx + bcy = e. Hence e/bd = (adx + bcy)/bd = (a/b)x + (c/d)y A. 7. Is it true that a subset of Q[2] (or of Z[2]) which is closed under substraction and squaring is also closed under multiplication? Answer: We did not know the answer to this question before the contest took place, but this did not prevent us from including it because our intension was to prepare a research oriented contest. Since then we know the answer for Z[2]: Yes, a subset of Z[2] which is closed under substraction and squaring is also closed under multiplication. Unfortunately the proof we have is not elementary (it is certainly above the highschool level) and needs some knowledge of intermediate level of abstract algebra. Let A be such a subset of Z[2]. We may assume that A . By question 1, A is closed under addition. It follows immediately that if n Z and A, then n A. Some intermediate abstract algebra implies that A = Z + Z for some , A. (Proof of this fact: Z[2] is a free Z-module freely generated by two elements (e.g. by 1 and 2). Thus A is also a Z-module generated by at most two elements. [Reference]). Thus every element of A can be written as n + m for some n, m Z. Since 2 A, we have 2 = u + v (1) for some u, v Z. Note also that A is closed under multiplication if and only if A. Assume v is odd. From (1) we get 22 = u2 + v. Thus v = 22 u2 (2) 2 2 2 Since and are in A, 2 is in A also. We also know that u A. It follows from (2) that v A. But v is odd and 2 A. By substracting (or adding) 2 from v as many times as needed we get A. We showed that if v is odd then is in A. Thus we may assume that v is even. Similarly we may assume that u is even. But if both u and v are even, then we can simplify (1) to get A. We still do not know the answer for Q[2]. 2 8. Ask your own questions of the same nature and try to answer them. 3