Thread: Uniqueness Proof

Hello everyone I'm practicing proofs and would like to know if I'm on the right track. Here's the problem:

Show that if x is a nonzero rational number, then there is a unique rational number y such that xy = 2

xy = 2
y = 2/x . . . solve for y

Since the equation is solved for y, this is the one and only value for y that makes the equation true. Further, since y is shown as a ration of 2 and x and x is a nonzero rational number, y is a rational number. Thus, y is a unique rational number.

Is this proof correct, or am I missing something?

Originally Posted by nicnicman

Hello everyone I'm practicing proofs and would like to know if I'm on the right track. Here's the problem:

Show that if x is a nonzero rational number, then there is a unique rational number y such that xy = 2

xy = 2
y = 2/x . . . solve for y

Since the equation is solved for y, this is the one and only value for y that makes the equation true. Further, since y is shown as a ration of 2 and x and x is a nonzero rational number, y is a rational number. Thus, y is a unique rational number.

Is this proof correct, or am I missing something?

A common way to show uniqueness is to suppose that there is more than one solution and show they were actually the same.

Suppose that there are two y values that make the equation true.

amd

Since both equations are true we can subtract them from each other, this gives



Since we are told that x is a non zero rational number the 2nd factor must be zero!



So there must have only been one solution after all!

Okay, I see that, but was my proof correct? Also, does your proof show that y is rational?

Originally Posted by nicnicman

Okay, I see that, but was my proof correct? Also, does your proof show that y is rational?

Your proof does not show it is unique without more. All you have proven is that there exists at least one solution.

No, my proof does not show it is rational. Now that it is unqiue you can use your method above to show what it must be.

Recall since

You have the equation



Now you need to argue why is a rational number. Be specific and cite any properties of the integers that you use.

2b/a is a rational number because x is expressed as ratio of 2b and a. Also, x is defined as a nonzero rational number. Therefore, a could not be zero. Thus, 2b/a is a rational number.

How is this?

As you said though, couldn't I have just used my previous proof to show that y is rational and avoided this last proof?

Originally Posted by nicnicman

2b/a is a rational number because x is expressed as ratio of 2b and a. Also, x is defined as a nonzero rational number. Therefore, a could not be zero. Thus, 2b/a is a rational number.

How is this?

As you said though, couldn't I have just used my previous proof to show that y is rational and avoided this last proof?

How do you know that ?

Because x is a nonzero rational number and x = a/b, b can not be 0. So 2b can not be 0.

I should also mention that we haven't got into sets yet.

Let me try this proof again:

Show that if x is a nonzero rational number, then there is a unique rational number y such that xy = 2

Solution:
Existence: The nonzero rational number y = 2/x is a solution of xy = 2 because x(2/x) = 2 = x(2/x) - 2 = x - x = 0.

Uniqueness: Suppose s is a nonzero rational number such that xs = 2. Then, xy =2 = xy - 2 = 0 and xs = 2 = xs - 2 = 0. Then:

xy - 2 = xs - 2
xy = xs
y = s

This would be a complete proof wouldn't it?

nicnicman, understand that the notion of 1/x or m/x is itself derived from uniqueness of y such that xy = 1 (or xy = m). So, what you are being asked is that one can use 1/x because it is unique. You cant use 1/x when proving a premise for its uniqueness. What you are doing here is proving the premise. The notions + and -, or * and / go hand in hand, one cant exist without another using unique numbers. Simply put, this proof of yours shows that 1/x can be written as a function of x, that is like, y = f(x) = 1/x. You cant use 1/x in this premise proof.

Salahuddin
Maths online

Thanks for the reply.

I don't understand how I used 1/x. Could you explain? Where did I go wrong? Is the existence proof correct?

See, there is nothing wrong in your "proof". But consider the following:

1. We know numbers, and thus we know 1/x and -x etc..
2. Enter group theory or vector algebra (These type of approaches and problems are usually in such fields of math).
1. Numbers do not exist, but sets under closed operations exist.
2. The elements of those sets have "similar" properties like numbers, like closure in addition/multiplication, commutativity, existence of an identity, and an inverse.
a. Notice also that, nothing is said that the "addition" on this set is like the "addition" on numbers, it is just notational similarity, may not be actual operational one.
3. These concepts are chained, as in, the idea of additive/multiplicative inverse depends on the idea of additive/multiplicative identity which in turn depends on the closure property.
4. Once you see this chaining, and the more abstract nature of the theory (than plain numbers and + and *), you will know why introducing 1/x before proving that "multiplicative inverse is unique" is frowned upon.

In other words, in such theories, 1/x is a short-hand for some function, that gives the (unique) multiplicative inverse of x. If we cant prove that such a function exists, there is no point in saying 1/x, because it does not exist, the notation just looks like we are taking reciprocal, but mind you, that is not the case. The x can be a vector or a modulo prime field element, for instance. The elegance of these group theory and vector algebras is that, they extend (using similar notation) the properties of numbers to much wider places.

So, your proof is right, but not to be used, because 1/x is not yet defined in the theory. Take your text book, and flip couple of pages forward. I am sure you will see a "definiition" of 1/x later, as the unique multiplicative inverse of x. Thats just notation, you know, it is not about just numbers.

Salahuddin
Maths online