The Gateway to Computer Science Excellence
0 votes
344 views
Hi Can anyone please explain this statements

S1: ∀x ∃y ∀z [ x+ y = z]

S2: ∃x ∀y ∃z [x + y = z]

Where x, y, z are real numbers. Which of the following statement is true?
in Mathematical Logic by (131 points) | 344 views

1 Answer

+3 votes
Best answer

S1: ∀x ∃y ∀z [ x+ y = z]
For all x, there exist y such that for all z. [x + y = z]. This is false.
Counter example: x = 2 and  y = 1 then x + y = 3. Then there is only one value of z that satisfy this. And that is z = 3. But here it is given for all z. So it should be true for z =1 , z = 2, z = 0. Clearly it is not true for all z.

S2: ∃x ∀y ∃z [x + y = z]
There exist x, such that for all y, there exist z which satisfy [x + y = z]. This is true.

Let take any x, x = 2 then for all y, there exist z which satisfy [y = z - 2]. This is true. Take what ever value of y there is always exist a z. which satisfy this. If y = 4 then z = 6. If y = 5 then z = 7.
What if i take x = 3, then also this is true.
So  ∀x ∀y ∃z [x + y = z] this is also true.

by Boss (16.5k points)
selected by
0
You are incorrect about the first problem, I think. It doesn't say 'for all y', it says 'there exists a y'. So for all x, there is some y such that for all z, x+y=z. Your choice of 1 for y was arbitrary, and not warranted given that it just says 'there exists a y', not that any y you choose will satisfy the condition. It's for all x and z, so let's choose 2 random numbers, say 3 and 34, for x and z. Now, is there some y that when added to 3 makes 34? Of course there is: 31. And that will hold true no matter what numbers you choose for x and z.

Although your answer to the second question was correct, the way you went about it shows the same error. There exists an x such that for all y, there exists a z such that x+y=z. You shouldn't start with any random x; although it works out in this case, that's the wrong approach. It's for all y, so y is the one where you can just choose any number, and the rest must be able to conform to it. So if you pick a 3 for y, you get x + 3 = z. It is easy to see that there will always be some combination of x and z that will make this true. But it is misleading to say 'let's choose a value for x' when it's 'there exists an x', not 'for all x'.

Incidentally, I'm assuming that the domain is the rationals. If it is the reals, then I'm not sure that either of these is true, since I don't think there is a specific real number that when added to pi, for instance, will make 5 (unless you're allowed to say pi plus (5 minus pi), of course).

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,397 answers
198,610 comments
105,454 users