Proof of a Biconditional Statement (p ↔ q)
Strategy
Prove that both p → q and its converse,
q → p, are true. That is, (Step 1) assume
p and deduce q, then (Step 2) assume q
and deduce p.
Special Case (Reversible proof)
If q can be deduced from p using only inferences of the form “iff”, then the argument
is called reversible and only one step is required to complete the proof.
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Proof of a Biconditional Statement (p ↔ q)
Theorem: An integer n is odd if and only
if n2 is odd.
Proof:
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Proof of a Biconditional Statement (p ↔ q)
Theorem:
x2 − 2x + 1 = 0 if and only if
x = 1.
Proof:
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Contraposition Proof of p → q
Theorem: If x2 − 2x − 3 > 0, then x < −1
or x > 3.
Proof:
Proof by Cases
Theorem: n2 −3n is even for all integers n.
Proof:
Proof by Contradiction (General Case)
Strategy
To prove a theorem of the form p by contradiction, we assume ¬p, then use rules of
inference to deduce a false statement (a
contradiction).
What this shows...
The fact that a valid argument produced a
false statement means that the premise ¬p
is false. Hence, ¬(¬p), which is equivalent
to p, is true.
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Proof by Contradiction (General Case)
Theorem:
There are no perfect squares
of the form 4k + 3, where k is an integer.
Proof:
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Proof by Contradiction (General Case)
Theorem:
For any real number x there
exists a positive integer n, such that x ≤ n.
Proof:
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Proof by Contradiction (General Case)
Theorem:
√
2 is an irrational number.
Proof:
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Proof by Contradiction (General Case)
Theorem:
There are an infinite number
of prime numbers.
Proof:
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Theorems of Equivalence
Some theorems state the equivalence of a
set of propositions {p1, p2, . . . , pn}.
Proof strategy
To prove pi ↔ pj for all i and j, we use an
n-step proof to establish a circular chain of
implications
Step 1: p1 → p2
Step 2: p2 → p3
...
Step n-1: pn−1 → pn
Step n: pn → p1
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