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Quantifiers

1.[2p]

What is the negation of ∀x>0(x2>x), over the reals?

Correct
The answer is: $\exists x > 0\, (x^2 \le x)$
The answer is: $\exists x > 0\, (x^2 \le x)$
The answer is: $\exists x > 0\, (x^2 \le x)$

2.[2p]

The claim "x2≥x for every x" is true in which of these domains?

Select all that apply

Correct
Correct
The answer is: The natural numbers $ℕ$, The integers $ℤ$
The answer is: The natural numbers $ℕ$, The integers $ℤ$

3.[1p]

Which symbolises "some prime is even", over the integers?

Correct
The answer is: $\exists n\, (n \ \text{is prime} \wedge n \ \text{is even})$
The answer is: $\exists n\, (n \ \text{is prime} \wedge n \ \text{is even})$
The answer is: $\exists n\, (n \ \text{is prime} \wedge n \ \text{is even})$

4.[1p]

A claim of the form ∀x(P(x)⇒Q(x)) is refuted by any value of x at which P(x) is false.

The answer is: False
Correct

5.[2p]

Over the integers, match each sentence to its symbolisation.

  • Every integer has a larger one

  • Some integer is larger than every integer

  • Every integer has a smaller one

  • Some integer is smaller than every integer

  • ∀x∃y(y<x)

  • ∀x∃y(y>x)

  • ∃y∀x(y<x)

  • ∃y∀x(y>x)

Show the answer

Every integer has a larger one: ∀x∃y(y>x) Some integer is larger than every integer: ∃y∀x(y>x) Every integer has a smaller one: ∀x∃y(y<x) Some integer is smaller than every integer: ∃y∀x(y<x)

6.[1p]

Over the integers, ∃y∀x(y>x) is true.

The answer is: False
Correct

7.[3p]

Which of these statements are true over the integers?

Select all that apply

Correct
Correct
Correct
Correct
The answer is: $\forall x\, \exists y\, (y > x)$, $\forall x\, \exists y\, (x + y = 0)$, $\exists y\, \forall x\, (x + y = x)$, $\exists y\, \forall x\, (xy = 0)$
The answer is: $\forall x\, \exists y\, (y > x)$, $\forall x\, \exists y\, (x + y = 0)$, $\exists y\, \forall x\, (x + y = x)$, $\exists y\, \forall x\, (xy = 0)$

8.[3p]

A sequence converges to L when ∀ε>0∃N∈ℕ∀n≥N(|an-L|<ε). Which statement says it does not converge to L?

Correct
The answer is: $\exists \varepsilon > 0\ \forall N \in ℕ\ \exists n \ge N\ (|a_n - L| \ge \varepsilon)$
The answer is: $\exists \varepsilon > 0\ \forall N \in ℕ\ \exists n \ge N\ (|a_n - L| \ge \varepsilon)$
The answer is: $\exists \varepsilon > 0\ \forall N \in ℕ\ \exists n \ge N\ (|a_n - L| \ge \varepsilon)$

9.[2p]

Put the lines of this proof that an=(-1)n does not converge to -1 in order.

  1. We prove the negation of the definition of convergence, with L=-1.

  2. Let N be any natural number, and take n=2N.

  3. Take ε=1.

  4. Then n≥N and n is even, so an=1 and |an-(-1)|=2≥1.

  5. Since N was arbitrary, (-1)n does not converge to -1.

Show the answer

a, b, c, d, e