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Proofs by Contraposition
o A proof by contraposition is based on the logical equivalence between
a statement and its contrapositive.
– Therefore, the implication p→ q can be proved by showing that its
contrapositive ~ q → ~ p is true. The contrapositive is usually proved directly.
o The method of proof by contrapositive may be summarized as:
1. Express the statement in the form if p then q.
2. Rewrite this statement in the contrapositive form
if not q then not p.
3. Prove the contrapositive by a direct proof.
Proofs by Contraposition
o Prove that for all integers n, if n2 is even then n is even.
PROOF: The contrapositive of the given statement is:
“if n is not even (odd) then n2 is not even (odd)”
We prove this contrapositive statement directly.
Suppose n is odd. Then n = 2k + 1 for some k ∈ Z
Now n2 = (2k+1) 2= 4k2 + 4k + 1
= 2•(2k2 + 2k) + 1
= 2•r + 1 where r = 2k2 + 2k ∈ Z
Hence n2 is odd. Thus the contrapositive statement is true and so the given
statement is true.
Proofs by Contraposition
For all integers m and n, if m + n is even then m and n are both even or m and n are both odd.
PROOF:
The contrapositive statement is:“For all integers m and n, if one of m and n is even and the
other is odd, then m + n is odd”
Suppose m is even and n is odd. Then
m = 2p for some integer p
and n = 2q + 1 for some integer q
Now m + n = (2p) + (2q + 1)
= 2•(p+q) + 1
= 2•r + 1 where r = p+q is an integer
Hence m + n is odd.
Similarly, taking m as odd and n even, we again arrive at the result that m + n is odd.
Thus, the contrapositive statement is true. Since an implication is logically equivalent to its
contrapositive so the given implication is true.
Proofs by Contraposition
Show that if 3n + 2 is an odd integer,then n is odd.
Proof : Assume that n is even.
This implies that n = 2k for some integer k. Then,
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1), so that 3n + 2
is even.
Since the negation of conclusion implies the
negation of hypothesis, the original conditional
statement is true.
Homework
1. Prove that if n2 is not divisible by 25, then n is not divisible by 5.
2. Prove that if n is an integer and n3 + 5 is odd, then n is even.
3. Prove that if 3n + 2 is odd, then n is odd

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MICROBIOLOGY biochemical test detailed.pptx
 
Alper Gobel In Media Res Media Component
Alper Gobel In Media Res Media ComponentAlper Gobel In Media Res Media Component
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Proofs by contraposition

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  • 2. Proofs by Contraposition o A proof by contraposition is based on the logical equivalence between a statement and its contrapositive. – Therefore, the implication p→ q can be proved by showing that its contrapositive ~ q → ~ p is true. The contrapositive is usually proved directly. o The method of proof by contrapositive may be summarized as: 1. Express the statement in the form if p then q. 2. Rewrite this statement in the contrapositive form if not q then not p. 3. Prove the contrapositive by a direct proof.
  • 3. Proofs by Contraposition o Prove that for all integers n, if n2 is even then n is even. PROOF: The contrapositive of the given statement is: “if n is not even (odd) then n2 is not even (odd)” We prove this contrapositive statement directly. Suppose n is odd. Then n = 2k + 1 for some k ∈ Z Now n2 = (2k+1) 2= 4k2 + 4k + 1 = 2•(2k2 + 2k) + 1 = 2•r + 1 where r = 2k2 + 2k ∈ Z Hence n2 is odd. Thus the contrapositive statement is true and so the given statement is true.
  • 4. Proofs by Contraposition For all integers m and n, if m + n is even then m and n are both even or m and n are both odd. PROOF: The contrapositive statement is:“For all integers m and n, if one of m and n is even and the other is odd, then m + n is odd” Suppose m is even and n is odd. Then m = 2p for some integer p and n = 2q + 1 for some integer q Now m + n = (2p) + (2q + 1) = 2•(p+q) + 1 = 2•r + 1 where r = p+q is an integer Hence m + n is odd. Similarly, taking m as odd and n even, we again arrive at the result that m + n is odd. Thus, the contrapositive statement is true. Since an implication is logically equivalent to its contrapositive so the given implication is true.
  • 5. Proofs by Contraposition Show that if 3n + 2 is an odd integer,then n is odd. Proof : Assume that n is even. This implies that n = 2k for some integer k. Then, 3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1), so that 3n + 2 is even. Since the negation of conclusion implies the negation of hypothesis, the original conditional statement is true.
  • 6. Homework 1. Prove that if n2 is not divisible by 25, then n is not divisible by 5. 2. Prove that if n is an integer and n3 + 5 is odd, then n is even. 3. Prove that if 3n + 2 is odd, then n is odd