How To Prove It
3 Proofs
3.1 Proof strategies
1. Consider the following theorem. (This theorem was proven in the introduction.)
Theorem. Suppose n is an integer larger than 1 and n is not prime. Then 2 n  1 is not
prime.
(a) Identity the hypotheses and conclusion of the theorem. Are the hypotheses true
when n  6 ? What does the theorem tell you in this instance? Is it right?
Hypotheses: n is an integer larger than 1 and n is not prime.
Conclusion: 2 n  1 is not prime.
When n  6, 2 6  1  63  32  7, which is not prime. For this particular instance, the
theory holds.
(b) What can you conclude from the theorem in the case n  15 ? Check directly that
this conclusion is correct.
When n  15, 215  1  32767  7  31  151, which is not prime. For this particular
instance, the theory holds.
(c) What can you conclude from the theorem in the case n  11?
Since 11 is a prime, we cannot conclude anything from this theorem.
2. Consider the following incorrect theorem:
Incorrect Theorem. Suppose n is a natural number larger than 2, and n is not a prime
number. Then 2n  13 is not a prime number.
What are the hypotheses and conclusion of this theorem? Show that the theorem is
incorrect by finding a counterexample.
Hypotheses: n is a natural number larger than 2, and n is not a prime number.
Conclusion: 2n  13 is not a prime number.
Counterexample: When n  8, which is not a prime 2n  13  2  8  13  16  13  29,
which is a prime, so this theorem does not hold.
3. Complete the following alternative proof of the theorem in Example 3.1.2.
Proof. Suppose 0  a  b. Then b  a  0.
[Fill in a proof of b 2  a 2  0 here.]
Since b 2  a 2  0, it follows that a 2  b 2 . Therefore if 0  a  b then a 2  b 2 .
(Proof)
Suppose 0  a  b. Then b  a  0.
Similarly b  a  0. Multiplying b  a to both sides of b  a  0, we have
(b  a )(b  a )  0(b  a)
b2  a2  0
Since b 2  a 2  0, it follows that a 2  b 2 . Therefore if 0  a  b then a 2  b 2 .
4. Suppose A \ B  C  D and x  A. Prove that if x  D then x  B.
(Preparation)
Contrapositive: If x  B then x  D.
(Proof)
Suppose A \ B  C  D. (1)
Suppose x  A and x  B. (2)
Hence, x  C and x  D.
Therefore, if A \ B  C  D, x  A, and x  B, then x  D, which implies if
A \ B  C  D, x  A, x  D then x  B.
5. Suppose a and b are real numbers. Prove that if a  b then
(Proof)
ab
 b.
2
Suppose a  b.
Adding b to both sides, we have
ab bb
a  b  2b
Dividing both sides by 2, we have
ab
 b.
2
Therefore, if a  b then
ab
 b.
2
6. Suppose x is a real number and x  0. Prove that if
(Preparation)
Contrapositive: If x  8 then
(Proof)
x 5 1
 .
x 6 x
3
2
Suppose x  8 .
3
x 5 3 8 5 25
7
1
 2


 .
Then 2
x  6 8  6 64  6 70 10
x 5 1
 then x  8.
x 6 x
3
2
On the other hand,
1 1
 .
x 8
Therefore, if x  8 then
3
x 5 1
x 5 1

,
 then x  8.
which
implies
if
2
2
x 6 x
x 6 x
3
7. Suppose a, b, c, and d are real numbers, 0  a  b, and d  0. Prove that if
ac  bd then c  d.
(Preparation)
Contrapositive: If c  d then ac  bd.
(Proof)
Suppose c  d. (1)
Suppose 0  a  b. (2)
Suppose d  0. (3)
Multiplying both sides of c  d (3) by a  0 (2) we have
ac  ad
Similarly, multiplying both sides of c  d (3) by b  0 (2) we have
bc  bd
Now, we will show ad  bd , i.e. bd  ad  0.
L.H.S. = bd  ad  d (b  a)
Since d  0 (3) and 0  a  b (2), which implies b  a  0,
 d (b  a )  bd  ad  0
Therefore, if c  d then ac  bd , which implies that if ac  bd then c  d.
8. Suppose x and y are real numbers, and 3x  2 y  5. Prove that if x  1then y  1.
(Preparation)
Contrapositive: If y  1 then x  1.
(Proof)
Suppose y  1. (1)
Suppose 3x  2 y  5. (2)
By (1), y  1, which implies 2 y  2.
So by (2), 3x  2  3x  2 y  5, which implies 3x  2  5 .
Hence,
3x  2  5
3x  3
x 1
Therefore, if y  1 then x  1, which implies that if x  1then y  1.
9. Prove the first theorem in Example 3.1.1. (Hint: You might find it useful to apply the
theorem from Example 3.1.2.)
Theorem 3.1.1. Suppose x  3 and y  2. Then x 2  2 y  5.
Example 3.1.2. Suppose a and b are real numbers. If 0  a  b then a 2  b 2 .
(Preparation)
Lemma: Suppose a  b and c  d. Then a  c  b  d.
(Proof to the lemma)
Suppose a  b. (1)
Suppose c  d. (2)
Adding c to both sides of (1), we have
a  c  b  c.
Similarly, adding d to both sides of (1), we have
a  d  b  d.
Now, we will show a  c  a  d.
By (2), a  c  a  d. is true.
Therefore, a  c  a  d  b  d , which implies
a  c  b  d.
(Proof to the exercise)
Suppose x  3. (1)
Suppose y  2. (2)
By (1), 0  3  x, and we can apply Example 3.1.2., which implies
32  x 2
9  x 2 (1)’
(2) implies 2 y  4 (2)’
By lemma, adding both sides of (1)’ and (2)’,
9  2y  x2  4
x2  4  9  2y
x2  2y  9  4  5
x2  2y  5
Therefore, if x  3 and y  2 then x 2  2 y  5.
10. Consider the following theorem.
2x  5
 3 then x  7.
x4
(a) What’s wrong with the following proof of the theorem?
2 x  5 2( 7 )  5 9

  3. Therefore if
Proof. Suppose x  7. Then
x4
74
3
2x  5
 3 then x  7.
x4
This “proof” violates the rule: “Never assert anything until you can justify it completely
using the hypotheses or using conclusions reached from them earlier in the proof.”
Theorem. Suppose x is a real number and x  4. If
(b) Give a correct proof of the theorem.
(Proof)
Suppose x  4. (1)
2x  5
 3. (2)
Suppose
x4
By (1) and (2),
2x  5
3
x4
2 x  5  3( x  4)
2 x  5  3 x  12
x7
Since x  7  4, this x value does not conflict the hypotheses.
2x  5
 3 then x  7.
Therefore, if
x4