COMP61 HW4 Solution Manual, prepared by Mamoon Raja Section 2.1 Question 2 Let us have a circle as shown in figure below. We divide the circle into 6 disjoint regions: ❖ The origin, by itself. (Region 1) ❖ All the points on the x­axis or y­axis that are not the origin, and lie within the boundaries of the circle. (Region 2) ❖ All the rest of the points within the boundary of the circle that are not on an axis, divided into 4 quadrants as shown in figure. (Regions 3­6) Let ‘m’ be the number of lattice points on the positive x axis. Then I claim there must also be m lattice points on each of the negative x axis, the positive y axis and the negative y axis, since the circle boundaries are equidistant from the origin. Thus the number of points in region 2 is of the form 4m. In addition, symmetry between the 4 quadrants gives that if one of the quadrants contains p points, then so do the others, so the number of points in regions 3­6 is of the form 4p. Finally the origin is 1 point. So the number of lattice points is 4m+4p+1, which is 4(m+p)+1 which is of the form 4k+1 for some positive integer k. Question 9. a. If ‘a’ is a multiple of 3, then we can write ‘a’ as a = 3x (Where, x ∈ Z) so, we can say that, a = x + x + x a = x + x + x + 1 ­ 1 a = ( x ­ 1 ) + (x) + ( x + 1 ) Thus, we can write ‘a’ as sum of three consecutive numbers b. Let’s start with a=4x. We will start with x=1 and check if we can write a as sum of four consecutive numbers. 0+1+2+3 = 6 ­1+0+1+2=2 => we can not write a as sum on 4 consecutive numbers. Let’s look at the case of x=2, here a=8 1+2+3+4 = 10 same conclusion for a=8, so we can say that the statement is not true. Question 13. Let us have a rational number ‘a’ and an irrational number ‘b’ where, a + b = c ………. (1) Let’s assume that c is a rational number, then we can write a and c as, a = m/n c = p/q so, eq.1 will become: mn + b = pq =>
=>
b = pq − mn b = pn − qm
qn As we can see that both ‘pn­qm’ and qn are integers, so b must be rational, but b is irrational. => Contradiction. Thus, sum of a rational and irrational number is irrational. Question 24. Additive identity element for all n ∈ Z exists because, n + 0 = n Now, we will prove that 0 is a unique additive identity. Let’s assume that we have two additive identity elements ‘a’ and ‘b’ for all n ∈ Z, where a ∈Z and b ∈ Z. Then, n + a = n ………... (1) n + b = n ……….... (2) Using eq.1 and eq., we can see that n + a = n + b =>
a = b Thus, any two additive identities for Z must be equal, so we can conclude that there is a unique additive identity element for Z.