31methods of Proof
31methods of Proof
31methods of Proof
(3,1)
Methods of Proof
1-Direct Proof
2- Proof by Contraposition
3- Proof by Contradiction
4- Proof by Cases
By:
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Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
Direct Proof: A direct proof shows that a conditional statement p q is true by showing that if p is true,
then q must also be true, so that the combination p true and q false never occurs. In a direct proof, we
assume that p is true and use axioms, definitions, and previously proven theorems, together with rules of
inference, to show that q must also be true
DEFINITION 1 The integer n is even if there exists an integer k such that n = 2k, and n is odd if there
exists an integer k such that n = 2k + 1. (Note that every integer is either even or odd,
and no integer is both even and odd.) Two integers have the same parity when both are
even or both are odd; they have opposite parity when one is even and the other is odd.
EXAMPLE 1 Give a direct proof of the theorem “If n is an odd integer, then is odd.”
Solution: n = 2k + 1 ,
is odd
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EXAMPLE 2 Give a direct proof that if m and n are both perfect squares, then nm is also a perfect square.
(An integer a is a perfect square if there is an integer b such that a = .)
Solution: We assume that m and n are both perfect squares. By the definition of a perfect square,
such that m = and n = . mn = =
. .
DEFINITION 3 The real number r is rational if there exist integers p and q with q 0 such that r = p/q.
p and q have no common factors (so that the fraction p/q is in lowest terms.)
A real number that is not rational is called irrational.
is rational number
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
DEFINITION 4 Let
EXAMPLE 4 : ,
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Exercises
1. Use a direct proof to show that the sum of two odd integers is even.
Solution:
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2. Use a direct proof to show that the sum of two even integers is even.
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
3. Show that the square of an even number is an even number using a direct proof.
Solution:
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4. Show that the additive inverse, or negative, of an even number is an even number using a
direct proof.
Solution:
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5. Prove that if m + n and n + p are even integers, where m, n, and p are integers, then
m + p is even.
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
6. Use a direct proof to show that the product of two odd numbers is odd.
Solution:
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7. Use a direct proof to show that every odd integer is the difference of two squares.
Solution:
( is odd)
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10. Use a direct proof to show that if is a prime number, then is a composite
number .
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
11. Use a direct proof to show that if is an odd prime number, then the number 4 is a
divisor to .
Solution:
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12. Use a direct proof to show that if is an odd number, then is odd.
Solution:
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Proof by Contraposition Proofs by contraposition make use of the fact that the conditional statement
p → q is equivalent to its contrapositive, ¬q →¬p. This means that the conditional statement p → q can
be proved by showing that its contrapositive, ¬q →¬p, is true. In a proof by contraposition of p → q, we
take ¬q as a premise, and using axioms, definitions, and previously proven theorems, together with rules
of inference, we show that ¬p must follow. the proof by contraposition can succeed when we
cannot easily find a direct proof.
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EXAMPLE 2 Prove that if n = ab, where a and b are positive integers, then
a≤ .
Solution:
Assume that
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is even .
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
Exercises
1. Use a proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers,
then x ≥ 1 or y ≥ 1.
Solution:
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2. Prove that if m and n are integers and mn is even, then m is even or n is even.
Solution:
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16. Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even .
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
17. Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd.
Solution:
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21. Use a proof by contraposition to show that if , where a and b are real
numbers, then or b 3.
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
22. Use a proof by contraposition to show that if x + y , where x and y are real
numbers, then x 4 or y 5.
Solution:
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Solution:
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Proofs by Contradiction Suppose we want to prove that a statement p is true. Furthermore, suppose
that we can find a contradiction q such that ¬p → q is true. Because q is false, but ¬p → q is true, we can
conclude that ¬p is false, which means that p is true. How can we find a contradiction q that might help us
prove that p is true in this way? Because the statement r ∧¬r is a contradiction whenever r is a
proposition, we can prove that p is true if we can show that ¬p → (r ∧¬r) is true for some proposition r.
Proofs of this type are called proofs by contradiction
is even
is even too
This is contradiction with the assumption that gcd( . is true .
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Exercises
1. Use a proof by contradiction to prove that the sum of an irrational number and a rational
number is irrational.
Solution:
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4. Use a proof by contradiction to show that there is no rational number r for which
[Hint: Assume that r = a/b is a root, where a and b are integers and a/b is in lowest terms.
Obtain an equation involving integers by multiplying by b3. Then look at whether a and b
are each odd or even.]
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
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6. Prove that is irrational by giving a proof by contradiction.
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
7. Let is irrational, prove that is irrational number.
Solution:
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8. Let is irrational, prove that is irrational number.
Solution:
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14. Use a proof by contradiction to prove that the product of an irrational number and a
non-zero rational number is irrational.
Solution:
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15. Let is a prime number . use a proof by contradiction to prove that if , then
.
Solution:
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17. Prove or disprove that the product of two irrational numbers is irrational.
Solution:
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18. Prove or disprove that the sum of two irrational numbers is irrational.
Solution:
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21. Prove or disprove that if then : .
Solution:
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
PROOF BY CASES A proof by cases must cover all possible cases that arise in a theorem.
We illustrate proof by cases with a couple of examples. In each example, you should check that
all possible cases are covered.
EXAMPLE 4 Use a proof by cases to show that |xy| = |x||y|, where x and y are real
numbers. ( Recall that |a| )
Solution: In our proof of this theorem, we remove absolute values using the fact that |a| = a
when a ≥ 0 and |a| = −a when a < 0. Because both |x| and |y| occur in our formula, we will
need four cases: (i) x and y both nonnegative,
(ii) x nonnegative and y is negative,
(iii) x negative and y nonnegative
(iv) x negative and y negative.
We denote by p1, p2, p3, and p4, the proposition stating the assumption for each of these
four cases, respectively.
(Note that we can remove the absolute value signs by making the appropriate choice of
signs within each case.)
Case (i):We see that p1 → q because xy ≥ 0 when x ≥ 0 and y ≥ 0,
so that |xy| = xy =|x||y|.
Case (ii): To see that p2 → q, note that if x ≥ 0 and y < 0, then xy ≤ 0,
so that |xy| =−xy = x(−y) = |x||y|. (Here, because y < 0, we have |y| = −y.)
Math 151 Discrete Mathematics [Methods of Proof] By: Malek Zein AL-Abidin
Case (iii): To see that p3 → q, we follow the same reasoning as the previous case with the
roles of x and y reversed.
Case (iv): To see that p4 → q, note that when x < 0 and y < 0, it follows that xy > 0.
Hence, |xy| = xy = (−x)(−y) = |x||y|.
Because |xy| = |x||y| holds in each of the four cases and these cases exhaust all possibilities,
we can conclude that |xy| = |x||y|, whenever x and y are real numbers.
EXAMPLE 7 Show that if x and y are integers and both xy and x + y are even,
then both x and y are even.
Solution: We will use proof by contraposition, the notion of without loss of generality, and
proof by cases. First, suppose that x and y are not both even. That is, assume that x is odd or
that y is odd (or both).Without loss of generality, we assume that x is odd, so that
x = 2m + 1 for some integer k. To complete the proof, we need to show that
xy is odd or x + y is odd. Consider two cases:
(i) y even
(ii) y odd.
In (i), y = 2n for some integer n, so that x + y =(2m + 1) + 2n = 2(m + n) + 1 is odd.
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