[Discrete Mathematics] Unique Quantifier Examples
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We evaluate the truth conditions of quantifiers and introduce the unique existential quantifier.
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Should there be a number subscript next to ∃ , so even when we need to say there exists 2 elements or whatever number ? if not is there a way to say there exists 2 numbers or 3 numbers, etc ?
Trivia: "∃!x Px" can be defined without the "!" by any of the following 2 equivalent formulas:
∃x [Px ∧ ∀y (Py → x = y)] :: There is something P, and anything P is that P :: There is x such that x is P and, for any y, if y is P, x is (identical to) y.
∃x Px ∧ ∀x∀y [(Px ∧ Py) → x = y] :: There is something P and if two things are P, they are actually the same thing :: There is x such that x is P and, for any x and y, if x is P and y is P, x is (identical to) y.
@5:50
The 2nd statement is incorrect as is the universe contains only one element. Then it is incorrect.
That's what I thought too
The (iii) exercise:
knowing that -1 times -1 = + 1
(-x)^2=(-x)(-x)=-1(x)-1(x)=+1(x)(x)=x^2
Whats the difference between 1question n 3question......??
Question on your example #3 on the first set of problems. Isn't x^2 different from (x)^2? such that -3^2 = -9 while (-3)^2 =+9, thus, (-(-3))^2 is not equal to -3^2?
x^2 and (x)^2 are the same thing. -x^2 and (-x)^2 are not.
I believe it would be false. Suppose we choose x = -3, this would give us (3)^2 and -3^2. One equals 9, the other is -9. This is assuming we aren't talking about the set of positive integers, you wrote x is an element of the set of integers, not only the positive ones.
how to prove ∃xp(x)→∃yq(y),∀x¬q(x)⊢∀x¬p(x) using quantifiers?
how to prove ∃xp(x)→∃yq(y),∀x¬q(x)⊢∀x¬p(x) using QN? thanks
did you change the thumbnail
hecc yee
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