00:01
So it looks like you have a few questions here about logic, and i'm happy to help.
00:05
So let's just start with number one.
00:08
For starters, i'll say that the only time that an implication is false is when we have true implies false.
00:15
So this statement right here is equal to a false.
00:19
And so my tactic here is that i'm hoping that i can eventually show that one side, say all of this, is equal to false, and this one is equal to true.
00:29
And i was thinking, well, how do i do that? well, how about i make this p implies q? what if i can make that true and have this r be false? and will that on the other side eventually give me an opposite answer? so this true implies false will be false.
00:51
And i'm hoping that with the same set of variables over here, i'll get true.
00:56
So let's just try this out.
00:57
Fortunately, we know from the problem statement that they are in fact not equivalent statements, and so i'm pretty sure that this is going to work.
01:08
So in order to make this p implies q to be true, i can do a few things.
01:15
I could set p equal to false, or i could set p equals to true as long as q is equal to true.
01:24
And i think i'll just choose the first one.
01:27
So i'll say p is equal to false.
01:30
And so on the right side, i'll get this p also being equal to false.
01:35
And then the next thing that i was looking at is that i want this r to be equal to a false as well.
01:44
Because i want my first statement to be true implies false.
01:48
And so this r should be false as well.
01:52
And looking on the right side, i'll just put an f there.
01:56
And fortunately, this is all the work i had to.
01:58
To do actually.
02:00
So if you notice here, the value for q can be really anything.
02:05
It could be either true or false and it doesn't really matter.
02:08
Because starting over here, if we have false implies anything, this will be true.
02:16
And then this r right here is false.
02:19
So true implies false.
02:21
And the left side will be an f.
02:23
Then on the right, we have the implication where the first statement is false.
02:28
And no matter what happens on the right side of an implication, where it starts out with false, we know that this should be true.
02:37
And so we have an f on the left side, a true on the right side.
02:42
And so this is one scenario that helps disprove or helps prove that these statements are not identical.
02:49
Next, let's go to 2a.
02:51
So we're supposed to determine the truth value of each statement, given some information already handed to us by the problem.
02:58
So 2a says, given not p or q is false, what's the true statement behind p implies q? so looking at this part right here, not p or q is equal to false.
03:13
We know that not p must be false and q must be false.
03:20
And so we can also rewrite that first one and say that p is true.
03:26
So we can plug in those values into our implication.
03:31
And so we'll have true implies false.
03:33
And as i stated earlier, this is the only scenario where an implication will give us a false value.
03:40
And so that's our answer for 2a.
03:43
For 2b, we're given that p and q is true.
03:48
And so we know that p must be true and q must be true for this to be possible.
03:53
And so then we can plug in a true on both sides of this double -in.
03:57
Now the way the double implication works is that it's a simultaneous p implies q and q implies p.
04:07
Now since both sides are true, then we can say that this double implication is true.
04:15
And you can verify it if you want to by going through a truth table with this as your final value.
04:22
P implies q and q implies p.
04:26
That's an equivalent statement to this double implication here.
04:33
All right, let's move on to part three.
04:36
Now, this one i wrote out in mathematical logic, but i'll just kind of read through it.
04:40
It says, there exists an x and y in the real numbers such that x and y are rational, and x plus y is irrational.
04:51
And now the proposed negation says, for all x and y in the real numbers, then the x plus y value is rational or sorry i'll say that one more time for all x and y in the real numbers if x plus y is rational then both x and y are both rational now we're supposed to determine whether these are negations and what'll do is i'll actually disprove both of these statements with examples so to disprove the first one we can just pick a certain set of numbers or here scratch that for the first one we can prove it just by using standard logic so if x and y are both rational numbers then x must be represented by some a over b and y must be represented by some c over d and all of these a bcd are integers and for space reasons i'm not going to write that out but i hope you can remember that then if we add them together we'll have x plus y is equal to a over b plus c over d and then this with a little bit of cross multiplication, just making sure that all of our variables are in the right spot.
06:06
We have ad plus cb over bd.
06:13
I also should have mentioned earlier that neither b nor d can be zero because then it wouldn't be a rational number.
06:23
So going back to this fraction right here, we know that a, b, c, and d are all integers, and so we have an integer over an integer and the denominator is not zero.
06:35
So x plus y is in fact a rational number.
06:39
So this line of logic disproves the first statement.
06:45
Now for the second statement, this is where i'll just pull out an example...