Solutions to Homework (week 2)

10 a, b and 14 were graded, 3 pts each, another 1 pt for completeness of other problems.

10 a. True.

Let n=0. Then for any (rational) number q, nq = 0 is an integer.

NOTE: Many people including me thought that the statement is false, because when n is not a 0, it is false. However, n=0 works.

If I graded this problem incorrectly (gave you less credits), please see me for a change.

10b True.

For any rational number q, by definition, q=a/b, where a, b are integers.

Then let n=b, nq = a is an integer.

14. Using the fact that the square of an odd is odd, the square of an even is even, we have

I, when a b both even, a^2 – b^2 is even

II, when a b both odd, a^2 – b^2 is even

III, when a even, b odd, a^2 – b^2 is odd

IV, when a odd, b even, a^2 – b^2 is odd

In summary, the sufficient and necessary condition for a^2 – b^2 to be odd is that one of a and b is odd, the other one is even.

21. Proof (using contradiction)

Suppose that the conclusion is false, i.e., a> sqrt(n) and b> sqrt(n). ß square root of n

The ab> n, which contradicts the assumption that n=ab. Q.E.D. ß end of the proof

NOTE: In general, there are 3 ways to show that A implies B or C,

Direct proof 1: Assume A is true and B is false. Need to show that C is true.

Direct proof 2: Assume A is true and C is false. Need to show that B is true.

By contradiction: Assume A is true, B and C are false. Then find a contradiction.