CSS342: Quantifiers. Professor: Munehiro Fukuda. Review of Propositions. Proposition: a statement that is either true or false, but not both Example: 1 < 4 is true. 2 > 5 is false. 3 is an odd number Then, how about x is an odd number ? The statement “X is an odd number”:
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CSS342: Quantifiers
Professor: Munehiro Fukuda
CSS342: Quantifiers
CSS342: Quantifiers
CSS342: Quantifiers
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CSS342: Quantifiers
is true.
x ≤ 1 and x > 1
(1 ) If x ≤ 1,
the hypothesis x > 1 is false,
thus the conditional proposition is true.
(2) If x > 1,
x + 1 > x (always true)
since x > 1, x + 1 > x > 1
Thus, the conditional proposition is true.
Therefore, this universally quantified statement is true.
CSS342: Quantifiers
is false.
x = 1
If x = 1, the proposition 12 – 1 > 0 is false
The value 1 is a counterexample.
Thus, this universally quantified statement is false
CSS342: Quantifiers
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CSS342: Quantifiers
is true.
x = 1
If n = 23,n + 1 = 24 (=3 * 8) is not prime
n + 2 = 25 (=5 * 5) is not prime
n + 3 = 26 (=2 * 13) is not prime
n + 4 = 27 (=3 * 9) is not prime
Thus, the proposition is true.
Therefor, this existentially quantified statement is true.
CSS342: Quantifiers
is false.
This in turn means that for all real number x, 1 / (x2 + 1) ≤ 1
Since 0 ≤ x2 for every real number x,
1 ≤ x2 + 1
By dividing both sides of this inequality expression by
x2 + 1
1 / (x2 + 1) ≤ 1
Is true.
Therefore, 1 / (x2 + 1) >1 is false for every real number x.
CSS342: Quantifiers
Domain D
Domain D
P(x) ≠ true(false)
P(x) = true
P(x) = true
≡
P(x) = true
P(x) = true
P(x) = true
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CSS342: Quantifiers
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Domain D
Domain D
P(x) = true
P(x) = false
P(x) = false
≡
P(x) = false
P(x) = false
P(x) = false
CSS342: Quantifiers
If P is a propositional funciton, each pair of propositions in (a) and (b) is logically equivalent, (i.e., has the same truth values.)
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CSS342: Quantifiers
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CSS342: Quantifiers
x, P(x)
we may prove x, P(x)
This in turn means that for all real number x,
1 / (x2 + 1) > 1 is false, (i.e., 1 / (x2 + 1) ≤ 1)
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Since 0 ≤ x2 for every real number x, 1 ≤ x2 + 1
By dividing both sides of this inequality expression by x2 + 1
1 / (x2 + 1) ≤ 1
CSS342: Quantifiers
Thus, x + y = x – x = 0
x, x + Y = 0
However, we can choose x = 1 – Y, such that
x + Y = 1 – Y + Y = 1 ≠ 0.
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CSS342: Quantifiers
You and your opponent play a logic game.
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CSS342: Quantifiers