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251 Notes 2.3 – Getting started with proofs
“... proofs are the only way to justify statements that are not obvious.” [p66]
Begin to prove or disprove by looking at examples
Do Exercise 1 on page 67. Then Exercise 2 on page 67.
To begin a proof, translate the problem using good suggestive notation.
Do Exercise 3 on page 68. Then Exercise 4 on page 68.
EXAMPLE 1 Prove that the product of two odd numbers is an odd number.
EXAMPLE 2 See Exercise 2(a): To prove a result, it is not sufficient to give some examples.
Two methods of proof:
Proof by cases
Proof by contradiction.
EXAMPLE 3. Proof by cases that n2 !2 is not divisible by 5 for any positive integer n:
(i) n = 5k
(ii) n = 5k+1 (iii) n = 5k+2 (iv) n=5k+3 (v) n= 5k+4
Compute each case and show the result is not a multiple of 5.
EXAMPLE 4. Do all the cases at once by n = 5k + r, where r is 0, 1, 2, 3, or 4 and show that
r2 ! 2 is either -2, -1, 2, 7, or 14 (none of which are divisible by 5).
If time permits, do exercises 5, 6, & 7 on page 70.
EXAMPLE 5. Proof by contradiction that there are infinitely many primes.
Consider Exercise 8 on page 71.
Exercises 2.3 p67.
1. For each false statement, provide a counterexample; for each statement that seems true, give supporting
examples.
a) The sum of three consecutive integers is always divisible by 3. 4+5+6=15=3A5; 8+9+10=27=3A9
b) The sum of four consecutive integers is always divisible by 4. NOT! 5+6+7+8=26 is not 4Ak
c) The sum of five consecutive integers is always divisible by 5. 2+3+4+5+6=20=5A4; 6+7+8+9+10=5A8
2. Provide a counterexample or give supporting examples.
a) If the product of positive integers m and n is 106, then 10 divides either m or n. NOT! m=26 and n=56
b) If p is prime, then so is p2 – 2. NOT! 112 – 2 = 119 = 7A17; 172 – 2 = 287 = 7A41
p68.
3. Which of the following are reasonable beginnings of a proof of a statement about two positive odd
integers?
a) Let m and n be positive odd integers. Then m can be written as 2k+1 and n can be written as 2j+1 for
nonnegative integers k and j. [Good beginning, although choice of letters might be i and j, or k and l.]
b) The odd integers can be written as 2k+1 and 2j+1 for nonnegative integers k and j. [Fine, unless it is
necessary to refer to the integers by name.]
c) Let m and n be positive odd integers. Then m can be written as 2k+1 for a nonnegative integer k. For the
same reason n can be written as 2k+1. [No! This is unacceptable where m and n may be different integers.]
d) Let m and n be positive odd integers. Then m = 2j+1 and n = 2k+1 for nonnegative integers j and k. [Fine.
This is essentially the same as part a).]
e) The odd integers can be written as 2x+1 and 2y+1 for nonnegative integers x and y. [Okay, but the letters
x and y typically suggest real numbers, whereas i, j, k, l, m, n are usually used to represent integers.]
f) Some positive odd integers are 7, 17, 25, 39, 73. If we check the result for all ten possible pairs, then that
should be convincing. [No number of specific examples is sufficient to do a general proof.]
4. Which of the following are reasonable beginnings of a proof of a statement involving a rational number and
an irrational number?
a) Let x be a rational number, and let y be an irrational number. [Okay, so far; but it makes no distinction
between rationals and irrationals.]
b) Let x be a rational number, and let y be an irrational number. The number x can be written as p/q where p,
q are in Z and q $1. [Fine. It completes the sense of part a) and uses a positive denominator for x.]
c) Let p be a rational number, and let q be an irrational number. [Worse than part a) because it uses unusual
choices for real (rational or irrational) numbers.]
d) Consider a rational number p/q and an irrational number x. [Okay, but it is not quite as complete as b).]
e) Let x = p/q be a rational number, and let y … p/q be an irrational number. [Shows confusion in regard to
specifications for y. It only says y is different from x, and rather should clarify that y cannot be written as a
quotient of any pair of integers.]
Page 70
5. Prove that 3 does not divide n2 – 2 for integers n $1.
The division algorithm allows us to write n = 3k + r for r = 0, 1, or 2.
Then n2 – 2 = (3k + r)2 – 2 = 9k2 + 6kr + r2 – 2.
We only need to show that r2 – 2 is not a multiple of 3 for each value of r.
Those values give -2, -1, or 2, so we are done.
6. Prove that 2 does not divide n2 – 2 for odd integer n.
Let n = 2k + 1, then n2 – 2 = (2k + 1)2 – 2 = 4k2 + 4k + 1 – 2 which is clearly odd [not divisible by 2].
7. Use the method of Examples 3 and 4 and Exercise 5 to
try to prove that 7 does not divide n2 – 2 for integers n $1.
a) Write n = 7k + r for r = 0, 1, 2, 3, 4, 5, or 6.
Now n2 – 2 = (7k + r)2 – 2 = 49k2 + 14kr + r2 – 2.
We only need to show that r2 – 2 is not a multiple of 7 for each value of r.
Now, as in Exercise 5, we see that the result holds for r = 0, 1, and 2.
However r = 3 gives r2 – 2 = 7, so the desired result does not hold.
[Result also fails when r = 4.]
b) n = 7(1) + 3 = 10 gives n2 – 2 = 98 which equals 7(14).
c) n = 7(0)+3 = 3, or 7(2)+3 = 17 are odd number counterexamples. Also 7(1)+4 = 11 is counterexample.
P71.
8. Prove the sum of a rational number and an irrational number is always irrational.
Let x be a rational number, and let y be an irrational number. The number x can be written as p/q where p,
q are in Z and q …0. A direct proof that x + y is not rational would not be easy; so we use the indirect method
of proof by contradiction.
If we assume x + y = m/n, for m, n in Z and n …0, then
, a rational number. But
y is given as not rational. This contradiction means x + y cannot be rational.
9. Prove that the product of two even integers is a multiple of 4.
Let m = 2k and n = 2l for integers k and l, then mAn =(2k)A(2l) = 4(kAl), which is a multiple of 4.
10. Prove that the product of an even integer and an odd integer is even.
Let m = 2k and n = 2l+1 for integers k and l, then mAn =(2k)A(2l+1) = 2(2kAl+k), which is even.
11. Prove that 5 does not divide n2 – 3 for integers n $1.
Let n = 5k + r, for r in {0, 1, 2, 3, 4}. Then n2 – 3 = (5k + r)2 – 3 = 25k2 + 10k + r2 – 3. We check that r2–3
is not divisible by 5 for r in {0, 1, 2, 3, 4}, because the results are respectively -3, -2, 1, 6, 13.