Thread: Prove or Disprove

1. Prove or disprove: There does not exist a real number x such that for all real numbers y, xy=1.

Contradiction: Suppose yes. That is, suppose that there is a real number x such that for all real numbers y, x*y=1.
Let x be the real number 0 and let y=z for any real number z.

Then,

x*y= 0*z .........by substitution
= 0 ......... by algebra

There is a contradiction, because when x is the real number 0, x multiplied by y equals 0. Therefore, the original statement is in fact true. There does not exist a real number x such that for all real numbers y, xy=1. **end of proof**

The only concern w/ my proof: Isn't the statement a universal statement? I thought that you are not allowed to use numbers to PROVE a universal statement using actual numbers, but this is the only proof I was able to come up with. Also, isn't this prove more of a counterexample as opposed to a contradiction?

2. Also, isn't this prove more of a counterexample as opposed to a contradiction?
In your case, one counter-example is enough to invalidate the original statement (or more exactly, validate it by invalidating its opposite). Check out my post on the previous page.

3. Isn't my above proof technically a counter example since I am using a number instead of a variable?

Is it CORRECT to disprove a universal statement that says 'there does not exist such a number x, such that...." when there really DOES exist such a number x with the use of contradiction, where in the contradiction the number ZERO is used to help prove the point instead of a variable? (because I was told the only time you are allowed to use numbers in your example is when u disprove a universal statement using counterexample)

4. You haven't proven the statement. You have just proven it for $x = 0$. You need to find a counter-example working with $y$. Again, go to the last post of the previous page.

5. Originally Posted by matthayzon89
Prove or disprove: There does not exist a real number x such that for all real numbers y, xy=1.

Contradiction: Suppose yes. That is, suppose that there is a real number x such that for all real numbers y, x*y=1.
Let x be the real number 0 and let y=z for any real number z.

Then,

x*y= 0*z .........by substitution
= 0 ......... by algebra

There is a contradiction, because when x is the real number 0, x multiplied by y equals 0. Therefore, the original statement is in fact true. There does not exist a real number x such that for all real numbers y, xy=1. **end of proof**

The only concern w/ my proof: Isn't the statement a universal statement? I thought that you are not allowed to use numbers to PROVE a universal statement using actual numbers, but this is the only proof I was able to come up with. Also, isn't this prove more of a counterexample as opposed to a contradiction?
Well, the statement isn't saying that we must prove or disprove it for all real numbers.

You only need one counterexample to show that xy cannot equal 1 for all real numbers.

xy=1 for an "infinity" of real numbers!! It just isn't for zero. That, of course is the only one that shoots it down.
Seems unjust but there you go, zero is a slippery one.

When you say "universal statement", you need to realise the question is asking you
if xy=1 for all real numbers y.

The only way you can show it's not is to discover at least one counterexample,
since clearly all non-zero reals have reciprocals.

6. Well, the statement isn't saying that we must prove or disprove it for all real numbers.
Yes it does, it wants us to prove that no real number $x$ can satisfy the condition $xy = 1$ with any real $y$. So just saying that the number $x = 0$ does not satisfy the condition does not prove the statement.

Well that's how I understand it. I think we should make it clear what the statement is before wasting lots of time and space.

7. y=1, x=1
y=2, x=1/2
y=3, x=1/3
y=k, x=1/k

y=0, x has no value

8. Originally Posted by Bacterius
Yes it does, it wants us to prove that no real number $x$ can satisfy the condition $xy = 1$ with any real $y$. So just saying that the number $x = 0$ does not satisfy the condition does not prove the statement.

Well that's how I understand it. I think we should make it clear what the statement is before wasting lots of time and space.
You need to reread the statement,
you're misinterpreting it.

9. Originally Posted by Archie Meade
You need to reread the statement,
you're misinterpreting it.
Yes, I've reread it and I still understand it in the same way. Prove that for any $y \in \mathbb{R}$, there exists no real number $x$ such that $xy = 1$. How are you interpreting the statement ?

10. It is okay to set x=0, in some cases, when using contradiction, is this correct?

11. Originally Posted by Bacterius
Yes, I've reread it and I still understand it in the same way. Prove that for any $y \in \mathbb{R}$, there exists no real number $x$ such that $xy = 1$. How are you interpreting the statement ?
hi Bacterius,

I know you meant for any y, because your proof followed from that.

However the statement is

"There does not exist a real number x, such that for all real numbers y, xy=1".

12. Originally Posted by Archie Meade
hi Bacterius,

I know you meant for any y, because your proof followed from that.

However the statement is

"There does not exist a real number x, such that for all real numbers y, xy=1".
Yes, and that's the base of my proof isn't it ? It is the word "all" that allows for a counter-example. Or not ... To be honest I'm a bit lost in the thread, it's confusing ... Gonna take some fresh air

13. We could express it numerically.

For all real numbers y.....

If y=1, there exists x=1 such that xy=1

so in that case there does exist an x for that y such that xy=1

and that is the only x that causes xy=1

If y=2, there exists x=0.5, such that xy=1

and that's the only x that causes xy=1

If y=3.5, there exists x=1/3.5, such that xy=1

Hence for almost any y, there exists an x=1/y such that xy=1

The only value of y that violates this is y=0.

Hence, there is one and only one y, such that an x does not exist to result in xy=1.

14. Originally Posted by Archie Meade
We could express it numerically.

For all real numbers y.....

If y=1, there exists x=1 such that xy=1

so in that case there does exist an x for that y such that xy=1

and that is the only x that causes xy=1

If y=2, there exists x=0.5, such that xy=1

and that's the only x that causes xy=1

If y=3.5, there exists x=1/3.5, such that xy=1

Hence for almost any y, there exists an x=1/y such that xy=1

The only value of y that violates this is y=0.

Hence, there is one and only one y, such that an x does not exist to result in xy=1.

I think you're confusing things here and this was clear from one of your prior posts to the OP, where you told him that his proof was correct if we rule out zero. Well no, his proof is incorrect even if we restrict ourselves to real non-zero numbers.

This the original question posted: "There does not exist a real number x such that for all real numbers y, xy=1

Change to "...for all non-zero real numbers y, xy = 1" ,and still the claim is true. In logic writing, the claim is: $\sim\exists x \,\forall y\,P(x,y)$ , where $P(x,y)$ is the property $xy=1$ .

This is WAY different from : for any non-zero real y there exists a real x s.t. xy = 1", or in symbols $\forall y\,\exists x\,P(x,y)$ ...

The original claims is: there doesn't exists a fixed number x s.t. for ANY real number y (zero or not, it works both ways) we'll get xy=1 . The claim is true in both cases (zero or not)...

Tonio

15. That's what I thought too I was starting to doubt of myself

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