Prove or Disprove

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• Mar 20th 2010, 10:09 AM
matthayzon89
Can someone help me prove or disprove
I need to Prove or Disprove: There does not exist a real number x such that for all real numbers y, xy=1.

This is false. I am going to try to prove it using contradiction.

Proof:
Suppose not. That is, suppose that there does exist a real number x such that for all real numbers y, xy=1. Let x be any real number r. X=1/r. Let y be an real number r. Y=r/1

Then,

X*Y= (1/r)*(r/1) by substitution
= 1 By algebra.

This is a contradiction because there does exist a real number x such that for all real numbers y x*y=1. **end of proof**

Can some someone please let me know if this is a correct way to write it on paper? I am pretty confident, however I would like to make sure, is there anything I can add for correctness?

Thank You,
Matt H.
• Mar 20th 2010, 10:53 AM
tonio
Quote:

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

This is false. I am going to try to prove it using contradiction.

You mean: "this is true", i.e.: the claim is true, since there indeed is NOT such a number x

Proof:
Suppose not. That is, suppose that there does exist a real number x such that for all real numbers y, xy=1.

You mean "suppose yes: there is such a number..." . I think you got confused with so many negations.

Let x be any real number r. X=1/r.

So you define $\displaystyle x=r\,,\,X=\frac{1}{r}=\frac{1}{x}$ ... why to choose the same letter, capital and non-capital, twice? This may help to make things even more confusing.

Let y be an real number r. Y=r/1

And now you've defined so far $\displaystyle x=y=r\,,\,\,X=Y=\frac{1}{r}$ ...is this really what you meant?

Then,

X*Y= (1/r)*(r/1) by substitution
= 1 By algebra.

This is wrong: $\displaystyle XY=\frac{1}{r}\frac{1}{r}=\frac{1}{r^2}$ ...and that's pretty much there is to it.

This is a contradiction because there does exist a real number x such that for all real numbers y x*y=1. **end of proof**

There is no proof at all above of anything. Perhaps you could try the following:

Suppose there exists such a real number $\displaystyle x\,\,\,s.t.\,\,\,xy=1\,\,\forall\,y\in\mathbb{R}$ . Choose now $\displaystyle y=2\Longrightarrow$[tex]
$\displaystyle 2x=1\Longrightarrow x=\frac{1}{2}$ , and now choose say $\displaystyle y=3$ and get that x must equal two different numbers: contradiction and we're done.

Tonio

Can some someone please let me know if this is a correct way to write it on paper? I am pretty confident, however I would like to make sure, is there anything I can add for correctness?

Thank You,
Matt H.

.
• Mar 20th 2010, 11:48 AM
matthayzon89
Im confused with ur explanation.

Our teacher told us to disprove this using contradiction:
There does not exist a real number x such that for all real numbers y, xy=1.

So i guess I need to assume that the above statement is true and write it out as if it was FALSE (suppose that there does exist a number x such that for all real number y, xy=1) and then go from there, and find out that there is no such number so assuming the "false" statement is true, is incorrect and the original statement is correct. right?
• Mar 20th 2010, 11:57 AM
matthayzon89
Quote:

So you define http://www.mathhelpforum.com/math-he...adbc625a-1.gif ... why to choose the same letter, capital and non-capital, twice? This may help to make things even more confusing.

Let y be an real number r. Y=r/1

And now you've defined so far http://www.mathhelpforum.com/math-he...46d9512a-1.gif ...is this really what you meant?

Can you please explain what is going on here? maybe describe it in words? this part got me losttt:)
• Mar 20th 2010, 01:09 PM
Quote:

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

This is false. I am going to try to prove it using contradiction.

Proof:
Suppose the statement is not true. That is, suppose that there does exist a real number x such that for all real numbers y, xy=1. Let x be any real number 1/r. X=1/r. Let y be an real number r. Y=r/1

Then,

X*Y= (1/r)*(r/1) by substitution
= 1 By algebra.

This (the statement) is a contradiction because there does exist a real number x such that for all real numbers y x*y=1. **end of proof**

Can some someone please let me know if this is a correct way to write it on paper? I am pretty confident, however I would like to make sure, is there anything I can add for correctness?

Thank You,
Matt H.

That's ok for all non-zero real numbers.
What about the real number zero?
• Mar 20th 2010, 03:37 PM
tonio
Quote:

Originally Posted by matthayzon89
Can you please explain what is going on here? maybe describe it in words? this part got me losttt:)

Well, apparently you succeeded to confuse me (Angry): you chose twice the same number (r), and then you define X = 1/r and Y = r/1 (here I thought you wrote i/r again).... why, in the devilish name of Pythagoras, would anyone write r/1 and not simply r???! Beats me...(Punch)

Tonio
• Mar 20th 2010, 04:07 PM
Plato
As a monitor I have tried to stay out of this.
BUT any reasonable axiom set for the real numbers is explicit or implies that
$\displaystyle \left( {\forall x \in \Re } \right)\left[ {x \cdot 0 = 0} \right]$
That alone makes this a nonsense question.
• Mar 22nd 2010, 02:56 PM
matthayzon89
grrrrrrrr im still lost, I dont understand really...so is tonio correct?
• Mar 22nd 2010, 03:00 PM
Quote:

Originally Posted by matthayzon89
grrrrrrrr im still lost, I dont understand really...so is tonio correct?

You would have been correct, only for the real value zero spoiling the party.
• Mar 22nd 2010, 03:46 PM
matthayzon89
isnt devision by zero impossible? it would make sense to consider it for this problem, would it?

im soooo darn confused, can someone please start from scratch and let me know if im on the right track, I am not even sure if my statement is true or false at this point, at first I thought it was false, but idk now b/c of the 'zero' case.
• Mar 22nd 2010, 03:55 PM
Quote:

Originally Posted by matthayzon89
isnt devision by zero impossible? it would make sense to consider it for this problem, would it?

im soooo darn confused, can someone please start from scratch and let me know if im on the right track, I am not even sure if my statement is true or false at this point, at first I thought it was false, but idk now b/c of the 'zero' case.

(0)x=0, so no value can multiply by zero to give 1.

Hence, your original proof is fine for all non-zero reals.

For any real non-zero number x=r, there is the real number y=1/r, such that xy=1

since the multiplicative inverse of x is the reciprocal 1/x.

The only real number without a reciprocal is zero.
• Mar 22nd 2010, 04:19 PM
matthayzon89
that kind of clarifies....thank you! ...So, you think a proof like mine is reasonable for the problem? or what you go about a different route trying to solve such a problem?
• Mar 22nd 2010, 04:37 PM
matthayzon89
moved***
• Mar 22nd 2010, 04:46 PM
Quote:

Originally Posted by matthayzon89
that kind of clarifies....thank you! ...So, you think a proof like mine is reasonable for the problem? or what you go about a different route trying to solve such a problem?

You made a good effort at expressing your proof, which didn't work because of zero only.

However, your proof was valid for every other real number,
hence you can say that you are correct for non-zero reals.

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

This statement is true, as in the case of y=0, as there is no x such that (0)x=1.

The statement is false if we exclude zero, however.

As the statement does not exclude zero, the statement is in fact true.

you should now try it again, including y=0.

I would not recommend a contradiction-type proof,
as the only culprit here is zero.
It's simplest to express the zero case as validating the statement,
without taking recourse to expressing a contradiction.
• Mar 22nd 2010, 04:59 PM
Bacterius
You can do it in an even simpler way.

Just say $\displaystyle y = 1$ and $\displaystyle y' = 2$. Now we need to find a real $\displaystyle x$ such as :

$\displaystyle xy = 1$
$\displaystyle xy' = 1$

This can be written (substituting the chosen values for $\displaystyle y$ and $\displaystyle y'$) :

$\displaystyle 1x = 1$
$\displaystyle 2x = 1$

Solve for $\displaystyle x$ :

$\displaystyle x = 1$
$\displaystyle x = \frac{1}{2}$

But $\displaystyle 1 \neq \frac{1}{2}$, therefore we have a contradiction that leads to the conclusion that there exists no real number $\displaystyle x$ that multiplied to any real $\displaystyle y$ because if I choose $\displaystyle y = 1$ then $\displaystyle y = 2$ (which are real), the values of $\displaystyle x$ differ. I have found a counter-example which is enough to invalidate the opposite of the statement by contradiction (and thus validating the statement).

If you prefer, if I can find a situation where the statement fails with two reals, then it does not work for all reals (therefore what you are trying to invalidate is invalidated).
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