John Landry

Answered

2022-07-15

For each of the following cases, is

$\forall k\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k){\textstyle )}$

true, false or dependent on the value of P(k)?

a) $\forall n\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}P(n),$

b) $P(0)\wedge P(1),$

c) $\forall n\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}P(2n).$

$\forall k\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k){\textstyle )}$

true, false or dependent on the value of P(k)?

a) $\forall n\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}P(n),$

b) $P(0)\wedge P(1),$

c) $\forall n\in \mathbb{N}\phantom{\rule{thickmathspace}{0ex}}P(2n).$

Answer & Explanation

Raul Garrett

Expert

2022-07-16Added 14 answers

Step 1

First, notice that the sentence

$A\to B$

is true whenever B is true. We use this fact in parts (a) and (c) below.

$\begin{array}{}\text{(\#)}& \forall k\in N\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k){\textstyle )}\end{array}$

a) $\forall n\in N\phantom{\rule{thickmathspace}{0ex}}P(n)$

a) true (because for any natural number n, P(n) is true so P(2n) will also be true

Yes, the given statement (a) tells us that P(2k) is true regardless of quantification; thus, so is $P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k);$; thus, (#) is true.

b) $P(0)\wedge P(1)$

b) true

Step 2

No. Here are two possibilities that are consistent with the given statement (b):

- statement (a) is true, in which case (#) is true,

- P(2) is false, in which case (#) is false.

Thus, we have insufficient information to conclude whether (#) is true or false.

c) $\forall n\in N\phantom{\rule{thickmathspace}{0ex}}P(2n)$

c) true

Yes, the given statement (c) tells us that $P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k)$ is true regardless of quantification; thus, (#) is true.

First, notice that the sentence

$A\to B$

is true whenever B is true. We use this fact in parts (a) and (c) below.

$\begin{array}{}\text{(\#)}& \forall k\in N\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k){\textstyle )}\end{array}$

a) $\forall n\in N\phantom{\rule{thickmathspace}{0ex}}P(n)$

a) true (because for any natural number n, P(n) is true so P(2n) will also be true

Yes, the given statement (a) tells us that P(2k) is true regardless of quantification; thus, so is $P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k);$; thus, (#) is true.

b) $P(0)\wedge P(1)$

b) true

Step 2

No. Here are two possibilities that are consistent with the given statement (b):

- statement (a) is true, in which case (#) is true,

- P(2) is false, in which case (#) is false.

Thus, we have insufficient information to conclude whether (#) is true or false.

c) $\forall n\in N\phantom{\rule{thickmathspace}{0ex}}P(2n)$

c) true

Yes, the given statement (c) tells us that $P(k)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}P(2k)$ is true regardless of quantification; thus, (#) is true.

Most Popular Questions