ASSIGNMENT 2
See questions here.
Contents:
b. $7 \leq p < 12$
c. $5 < x < 7$
d. $x < 4$ (since $x < 4 \implies x < 6$)
e. $-3 < y < 3$ (note that $y^2 < 9 \implies y < 4$ and $y^2 < 9 \implies -3 < y < 3$)
f. $x = 0$
b.
$p$ ranges from 7 to below 12.
c.
$x$ lies between 5 and 7.
d.
$x$ is lesser than 4.
e.
$x$ lies between -3 and 3.
f.
$x$ equals zero.
To show $\phi_1 \land \phi_2 … \phi_n$ is true, it is necessary and sufficient to show that each of $\phi_1, \phi_2 … \phi_n$ is true.
To show $\phi_1 \land \phi_2 … \phi_n$ is false, it is sufficient to show that any one of $\phi_1, \phi_2 … \phi_n$ is false.
To help see the answer, observe the following:
| if $\pi > 3$ is… | then $\pi > 10$ is… |
|---|---|
| 0 | 0 |
| 1 | 0 or 1 |
In the second row, the truth value of $(\pi > 3) \lor (\pi > 10)$ is independent of the truth value of $\pi > 10$. The first row shows that there is no case where $\pi > 10$ is true if $\pi > 3$ is false.
| if $\pi > 10$ is… | then $\pi > 3$ is… |
|---|---|
| 0 | 0 or 1 |
| 1 | 1 |
In the first row, the truth value of $(\pi > 3) \lor (\pi > 10)$ is dependent on the truth value of $\pi > 3$. The second row shows that there is no case where $\pi > 3$ is false if $\pi > 10$ is true.
Already, we see that the truth value of $(\pi > 3) \lor (\pi > 10)$ is independent of the truth value of $\pi > 10$ and dependent entirely on the truth value of $\pi > 3$. To confirm this, see the following truth table:
| $\pi > 3$ | $\pi > 10$ | $(\pi > 3) \lor (\pi > 10)$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | NA |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Removing the NA row:
| $\pi > 3$ | $\pi > 10$ | $(\pi > 3) \lor (\pi > 10)$ |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Here, we see that the truth value of $(\pi > 3) \lor (\pi > 10)$ exactly corresponds to the truth value of $\pi > 3$. Hence, $(\pi > 3) \lor (\pi > 10)$ can be simplified to $\pi > 3$.
The only value $x$ cannot take is 0. Hence:
$(x > 0) \lor (x < 0) \equiv x \neq 0$
$x \geq 0$
$x \geq 0$ (the $x > 0$ clause is contained in this)
$x^2 > 9 \implies (x < -3) \lor (x > 3)$
Hence:
$(x > 3) \lor (x^2 > 9)$
$\equiv (x > 3) \lor (x < -3) \lor (x > 3)$
$\equiv (x < -3) \lor (x > 3)$
$\equiv x^2 > 9$
a.
$\pi$ is greater than 3.
b.
$x$ is not equal to 0.
c.
$x$ is greater than or equal to 0.
d.
$x$ is greater than or equal to 0.
e.
The square of $x$ is greater than 9.
To show $\phi_1 \lor \phi_2 … \phi_n$ is true, it is sufficient to show that any one of $\phi_1, \phi_2 … \phi_n$ is true.
To show $\phi_1 \lor \phi_2 … \phi_n$ is false, it is necessary and sufficient to show that each of $\phi_1, \phi_2 … \phi_n$ is false.
$\lnot (\pi > 3.2) \equiv \pi \leq 3.2$
$\lnot (x < 0) \equiv x \geq 0$
$x^2 > 0 \implies x \neq 0$
Hence:
$\lnot (x^2 > 0)$
$\equiv \lnot x \neq 0$
$\equiv x = 0$
$\lnot (x = 1)$
$\equiv (x < 1) \lor (x > 1)$
$\equiv x^2 > 1$
$\lnot \lnot \psi \equiv \psi$
a.
$\pi$ is less than or equal to 3.2.
b.
$x$ is greater than or equal to 0.
c.
$x$ is not equal to 0.
d.
The square of $x$ is greater than 1.
e.
$\psi$ holds true.
For reference:
$D =$ “Dollar is strong”
$Y =$ “Yuan is strong”
$T=$ “New US–China trade agreement signed”
“Dollar and Yuan are both strong.”
$D \land Y$ (simple conjunction)
“Trade agreement fails on news of weak Dollar.”
$\lnot T \land \lnot D$
The causal relationship between the news and the failure is irrelevant to the truth of the conjunction inherent in the statement, i.e. that the Dollar is not strong ($\lnot D$) and the new US-China trade agreement is not signed ($\lnot T$).
“Dollar weak but Yuan strong, following new trade agreement.”
$\lnot D \land Y \land T$
Again, causality is irrelevant to the inherent conjunction.
“Strong Dollar means a weak Yuan.”
$D \land \lnot Y$
“Yuan weak despite new trade agreement, but Dollar remains strong.”
$T \land \lnot Y \land D$
“Dollar and Yuan can’t both be strong at same time.”
$(D \land \lnot Y) \lor (\lnot D \land Y)$
“If new trade agreement is signed, Dollar and Yuan can’t both remain strong.”
$T \implies \lnot (D \land Y)$
“New trade agreement does not prevent fall in Dollar and Yuan.”
$T \not \implies \lnot (D \land Y)$