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zolvtas

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Feb 7, 2021
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Hi.

Would appreciate a lot if someone helped me with the following problem:

Problem
Let A = {0,1,2,3,4}.
Is the following expression true or false: ∀x∈A, ∃y∈A: x+y<5

Correct answer
The correct answer is true which I dont understand. From what I can tell the expression says "For all x that belongs to A, there exist y that belongs to A", but that would make the expression false in my opinion.

Thanks
Hann
 
Last edited:
zolvtas said:
Hi.

Would appreciate a lot if someone helped me with the following problem:

Problem
Let A = {0,1,2,3,4}.
Is the following expression true or false: ∀x∈A, ∃y∈A: x+y<5

Correct answer
The correct answer is true which I dont understand. From what I can tell the expression says "For all x that belongs to A, there exist y that belongs to A", but that would make the expression false in my opinion.

Thanks
Hann
Click to expand...
Not quite. It means, "For all x that belongs to A, there exist y that belongs to A such that x+y<5"

But even taking your interpretation, why would that be false? It would say that, whatever element you pick, I can pick an element. Surely I could?

When you write back, be sure to explain why you make your conclusion.
 
Thanks for quick reply!

Lets start over..

My first attempt to understand this problem is to look for some logic between the numbers in the brackets and the expression. For example by assuming the numbers in the brackets are all x-values, and for every x vaule there is an y value that is the exact same. This means that x1=0 gives y1=0 as well as x4=0 give y4=0. Since x4=4 and y4=4 is more than 5 I though that the this expression was false.

Do I make any sense at all to you?
 
zolvtas said:
My first attempt to understand this problem is to look for some logic between the numbers in the brackets and the expression. For example by assuming the numbers in the brackets are all x-values, and for every x vaule there is an y value that is the exact same. This means that x1=0 gives y1=0 as well as x4=0 give y4=0. Since x4=4 and y4=4 is more than 5 I though that the this expression was false. Do I make any sense at all to you
Click to expand...
What you say is clear, showing that you do not understand the difference between \((\forall x\in A)\) and \((\exists y\in A))\)
The first says that for all elements in \(A\) there is some element in \(A\)) and when added together give a sum less than five.
Note the every(all) and the some(at least one).
\((\forall x\in A)(\exists y\in A)[x+y<5]\) translates to every term of \(A\) there is a term of \(A\) so that their sum is less that five.
 
Thanks for support!

Do I understand you correct? -->
(∀x∈A) and (∃y∈A))[x+y<5] = to every term of A there is a term of A so that their sum is less that five.
(∀x∈A) and (∃y∈A)= For all elements in A there is some element in A.

Also:
Which expression means some?
What role do x and y have?
Can you show what these expressions have to do with "Let A = {0,1,2,3,4}."?
 
zolvtas said:
My first attempt to understand this problem is to look for some logic between the numbers in the brackets and the expression. For example by assuming the numbers in the brackets are all x-values, and for every x vaule there is an y value that is the exact same. This means that x1=0 gives y1=0 as well as x4=0 give y4=0. Since x4=4 and y4=4 is more than 5 I though that the this expression was false.
Click to expand...
No, you don't need to "look for some logic"; you need to read the logic they have stated.

No, the numbers in the given set A = {0,1,2,3,4} are not "x-values"; they are numbers in a set, which one may assign to any variable at will. In fact, values of both x and y here may come from set A -- and they can be chosen independently, which seems to be your error.

The expression "∀x∈A, ∃y∈A: x+y<5" does say "for every x value there is a y value ...", but not "that is the exact same". That would require it to say x=y. (If it did say that, the statement would not be false, though. Given any value of x, you can surely choose to set y to the same value, so such a value exists, and the statement is true.)

What it says is, "for every x value there is a y value such that their sum is less than 5". This means, for example, if I choose to set x=0, then you can find a value of y such that 0+y<4; you might, for example, choose y=1. Go through each possible choice for x and see whether you can always make the sum less than 4, because that is what the statement claims. What value might you choose for x that will make it impossible? When you find that, then you can say that the statement is false.
 
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