# Uniqueness Proof

#### 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?

#### TheEmptySet

MHF Hall of Honor
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.

$$\displaystyle xy_1=2$$ amd $$\displaystyle xy_2=2$$

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

$$\displaystyle xy_1-xy_2=0 \iff x(y_1-y_2)=0$$

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

$$\displaystyle y_1-y_2=0 \iff y_1=y_2$$

So there must have only been one solution after all!

#### nicnicman

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

#### TheEmptySet

MHF Hall of Honor
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 $$\displaystyle x \in \mathbb{Q} \implies x=\frac{a}{b}, a,b \in \mathbb{Z} \quad b \ne 0$$

You have the equation

$$\displaystyle \frac{a}{b}\cdot y =2 \iff y=\frac{2b}{a}$$

Now you need to argue why $$\displaystyle \frac{2b}{a}$$ is a rational number. Be specific and cite any properties of the integers that you use.

#### 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?

#### TheEmptySet

MHF Hall of Honor
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 $$\displaystyle 2\cdot b \in \mathbb{Z}$$?

#### nicnicman

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.

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#### nicnicman

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?

#### Salahuddin559

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

#### nicnicman

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?

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