Prove that there exists an ideal integer.
Since the proper divisors of 6 are 1, 2 and 3 and 6 = 1 + 2 + 3, 6 is an ideal number.
The purpose of this question was to see if you could apply a simple definition and whether you could do a constructive existence proof.
Using the definition of an odd integer, prove that, for every positive integer n, if n is odd then n is the difference of two squares. (That is, n = i2 − j2 for two integers i and j.)
(Hint. Look at (k+1)2 − k2.)
Suppose n is an odd positive integer. Then n = 2k + 1 for some integer k. Notice that (k + 1)2 − k2 = 2k + 1. So n = (k + 1)2 − k2, showing that n is the difference of two squares.
The purpose of this question was to see whether you can carry out a simple direct proof of an implication.
Using the definition of the harmonic mean of two numbers, prove by using the contrapositive that, for every two positive real numbers x and y, if x < y then the harmonic mean of x and y is less than y.
The contrapositive of the statement is: If the harmonic mean of x and y is ≥ y then x ≥ y. We prove that using a direct proof.
Suppose that the harmonic mean of x and y is ≥ y. That is,
Since 1/x + 1/y = (y + x)/xy and x and y are both positive,
| 2xy/(x + y) ≥ y |
| → 2x/(x + y) ≥ 1 |
| → 2x ≥ x + y |
| → x ≥ y |
The purpose of this question was to see whether you can carry out a proof of an implication using the contrapositive.
Prove or disprove the following statement. If it is false, give a counterexample.
For all positive integers n and m, if nm + n is odd then either n is even or m is odd.
The statement is false. n = 1 and m = 2 is a counterexample, since then nm + n is odd, n is odd and m is even.
The purpose of this question was to see if you think before you start to write a proof.