Question

Hi, I am having a problem determining whether or not the statement is true. (Help would be nice.)

Answer 1

Hello! what's your statement?

Answer 2

\(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\)

Answer 3

@jim_thompson5910 you can have this one, buddy.

Answer 4

I know how to do \(\color{blue}{\displaystyle(∃x)(∃!y)(x^2-y^2=64)}\).
When x=8 (or x=-8), leaving me with only one option, y=0.

So that would have been true.

Answer 5

The way I read \(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\) is,
"there exists a unique (the only) value of y, so that there will be (will exist) an x-value to make the statement true"

and I think that this is in fact not true

Answer 6

Because, indeed \(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64)}\)

(for all y, not just for a unique value of y)

Answer 7

(I will have to leave for about 30 minutes sorry.)

Answer 8

I think the commutative law applies in this case

https://www.cs.sfu.ca/~ggbaker/zju/math/pred-quant.html

scroll to the bottom to the "Nested Quantifiers" section

Answer 9

Focus on where it says "But it turns out these are equivalent:...i.e. you can swap the same kind of quantifier"

Answer 10

if x=10,y=6
can you check?

Answer 11

100 - 36 = 64. ^

Answer 12

I believe the order does count in relevance to the x and y values though.

Answer 13

i have tried like this .
(x+y)(x-y)=64
both can be negative.

Answer 14

either x=10,y=6
or x=-10,y=-6

Answer 15

Yes, which both work in this situation. But, I believe a certain unique value must be found, not completely sure.

Answer 16

@idku May you clarify the problem for me a little more? It's been awhile since I touched base on these.

Answer 17

@sshayer , so basically you are saying that I can define what my unique valye of y is?

Answer 18

As x = 8; x = -8, y must equal 0.
In relevance toward, 10 and 6, you are able to configure a problem in which both values in positive and negative terms allow for both formats to succeed in the answer of 64.
You may define your unique value of y.

Answer 19

In the statement: \(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\)
I think you are first to choose the y, and then the x.

Answer 20

Yes, in sense of the unique value. Y before the x.

Answer 21

I may be incorrect, or in my mind sensing the x may be first, so I will not hold my statement to be completely true.

Answer 22

Basically the way I read \(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\) is:

"There is only one value of \(\color{blue}{\displaystyle y}\), for which there will
exist (at least one value of) \(\color{blue}{\displaystyle x}\) to satisfy \(\color{blue}{\displaystyle x^2-y^2=64}\)."

Answer 23

(If "∃x" was first, it would be a lot easier, but I think these aren't equivalent, and I just explained why I think so.)

Answer 24

Oh, I see now. Yes, I would believe you must choice the y value first, then the x value after.

Answer 25

Really, the following is true: \(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64)}\)

So, if it is true \(\forall y\), how can I say that there exists a unique value of y (only one value for y) that would satisfy the statement.

Answer 26

I apologize, I have forgot much about this, but I can surely inform when y = 0 as you stated as the unique value for y, does allow for at least one value of x to construct toward 64. I wish I could be of further assistance. Maybe @jim_thompson5910 could help you with this part in which you state there exists a unique value of y.

Answer 27

Yeah, I am pretty certain the preposition is false.

What does \(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\) say?

There will exist ONLY one value of \(y\), for which there
will exist AT LEAST ONE \(x\) to satisfy \(x^2-y^2=64\).

This is not true tho', because FOR ANY VALUE OF \(y\) ((not for a unique one)) ,
there will exist an \(x\) to satisfy \(x^2-y^2=64\).

Answer 28

This is basically the argument I made (just want confirmation, or dispute).

Answer 29

Confirmed for one which is not a unique value of y, I would believe so, yes. The preposition would be considered false in my eyes.

Answer 30

I will repost my basic argument to save time.

Answer 31

Ignore my post above. It only works for existential quantifiers (not unique existential quantifiers)

You are correct @idku the statement would be false. Simply list 2 or more cases and that will be enough to prove the uniqueness aspect doesn't hold up.

Answer 32

The proposition: \(\color{blue}{\displaystyle(∃!y)(∃x)(x^2-y^2=64)}\)

My interpretation:

There exists a unique \(y\), for which there is going to
exist (at least one) \(x\), to satisfy \(x^2-y^2=64\).

Answer 33

tnx for the confirmation, Jim.

Answer 34

So, in any case, I claimed that in fact
\(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64)}\) is true, and
it is certainly not a unique value of \(y\)
for which the rest of the statement holds.

Answer 35

You don't need to show it for all y

You just need more than one y. That's the min amount of work you need to do really.

Answer 36

But that method will work too

Answer 37

y = 0, y = +/- 6, y = +/- 15 are some choices to equal 64 as the result.
x = +/- 8, x = +/- 10, x = +/- 17.

Answer 38

So, if I actually want to disproof, I just need to specific (numerical) examples?

Answer 39

need two**

Answer 40

I believe so, as what Jim stated.

Answer 41

I gave you 3 cases, just so you could see there was more.

Answer 42

`So, if I actually want to disproof, I just need two specific (numerical) examples?`
correct

Answer 43

To me, if your statement is A, then if \(\neg A\) is False, then A is true.
And \(\neg A \) is \(\forall y , \cancel \exists x | x^2-y^2=64\)
this statement is False, hence A is true.

Answer 44

I lost you on how "for all y" is really a negation to "for a unique y".
There are multiple possibilities (also "exists no y such that....)

Answer 45

The original one is "There exists only y", negate it gives you "for all y"

Answer 46

And actually, Loser66, I think that \(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64)}\) would be true.

Answer 47

If you play guess game, then go with it
If you need a proof, I think my way works well. :)

Answer 48

For any value of \(y\), there will exist an \(x\) to satisfy \(x^2-y^2=64\).

It's an equivalent of
\(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64) ~~:= ~~}\)\(\color{blue}{\displaystyle(\forall y)(∃x)(|y|=\sqrt{x^2-64})}\)

Answer 49

You said that \(\color{blue}{\displaystyle(\forall y)(∃x)(x^2-y^2=64)}\) would not be true, can you show me how then? (And no, I am not guessing, just trying to learn something here:D)

Answer 50

Well, he left, I guess I will go with my conclusion of the initial statement being false (and will provide two examples of this).

Answer 51

Thanks for participation ;;

Answer 52

My statement is "for all y" "NOT exist x" such that x^2-y^2=64
That is wrong.

Answer 53

The only time \[\color{blue}{\displaystyle(\exists ! y)(\exists x)(f(x,y)=0)}\]
is true is if you have a horizontal line for the graph of f(x,y)=0. However, x^2-y^2 = 64 leads to a hyperbola. So that's another way to look at it.