Proving that the inverse of a rational number exists

[brotherbobby]

Homework Statement

Let $a=\frac{m}{n}$ be a rational number expressed as a quotient of integers $m,n$ with $m\ne 0\;\text{and}\; n\ne 0$. Show that there is a rational number $b$ such that $\boldsymbol{ab=ba=1}$

Relevant Equations

None, as far as I can see.

Problem statement : I cope and paste the problem as it appears in the text below.

Attempt : Not being a math student, I try and prove the above statement using an "intuitive" way.

Let us have a rational number $b = \frac{n}{m}$. Multiplying with $a$ from the right, we see $ab = \frac{m}{n}\frac{n}{m} = 1$. Since it is given that $m,n\ne 0$, we can see that a $b=\frac{n}{m}$ will always exist.

Likewise, we can multiply $a$ from the left by $b$ and show that $b$ exists.

This proves the theorem.

Issue : Am I correct with the reasoning above? A hint or correction would be welcome.

 

[fresh_42]

Homework Statement:: Let $a=\frac{m}{n}$ be a rational number expressed as a quotient of integers $m,n$ with $m\ne 0\;\text{and}\; n\ne 0$. Show that there is a rational number $b$ such that $\boldsymbol{ab=ba=1}$
Relevant Equations:: None, as far as I can see.

Problem statement : I cope and paste the problem as it appears in the text below.

View attachment 314974

Attempt : Not being a math student, I try and prove the above statement using an "intuitive" way.

Let us have a rational number $b = \frac{n}{m}$. Multiplying with $a$ from the right, we see $ab = \frac{m}{n}\frac{n}{m} = 1$. Since it is given that $m,n\ne 0$, we can see that a $b=\frac{n}{m}$ will always exist.

Likewise, we can multiply $a$ from the left by $b$ and show that $b$ exists.

This proves the theorem.

Issue : Am I correct with the reasoning above? A hint or correction would be welcome.

Sure, you are right.

The question here is: What is a rational number, and what can we use to prove the statement?
If you say: $b=\frac{n}{m}$ does the job, then you implicitly claim that this is a rational number, and that there is a multiplication that allows you to build $\frac{m}{n} \cdot \frac{n}{m}$ and get $1$ as a result.

For example, I prefer to consider rational numbers as pairs $(n,m)$ of integers where $m\neq 0$ and define the equality of two rational numbers by $(n,m)=(n',m')\Longleftrightarrow n\cdot m' = m\cdot n'.$ Multiplication and addition have to be defined accordingly. Whatever, it provides an environment that allows us to apply rules and decide equality. Without such an environment we will be left with "Sure. What else?"

 

[brotherbobby]

Sure, you are right.

The question here is: What is a rational number, and what can we use to prove the statement?
If you say: $b=\frac{n}{m}$ does the job, then you implicitly claim that this is a rational number, and that there is a multiplication that allows you to build $\frac{m}{n} \cdot \frac{n}{m}$ and get $1$ as a result.

For example, I prefer to consider rational numbers as pairs $(n,m)$ of integers where $m\neq 0$ and define the equality of two rational numbers by $(n,m)=(n',m')\Longleftrightarrow n\cdot m' = m\cdot n'.$ Multiplication and addition have to be defined accordingly. Whatever, it provides an environment that allows us to apply rules and decide equality. Without such an environment we will be left with "Sure. What else?"

Yes, however I must admit, perhaps because I am not a Math student, that I did not understand what is asked. Or perhaps it looks so obvious that I don't know what am I supposed to show.

I know that in Math the use of "necessary" and "sufficient" is vital. Can the question above be posed in a way where the existence of the inverse of a rational number $a$ such that $ab=ba=1$ is both necessary and sufficient?

 

[fresh_42]

Yes, however I must admit, perhaps because I am not a Math student, that I did not understand what is asked. Or perhaps it looks so obvious that I don't know what am I supposed to show.

I know that in Math the use of "necessary" and "sufficient" is vital. Can the question above be posed in a way where the existence of the inverse of a rational number $a$ such that $ab=ba=1$ is both necessary and sufficient?

As mentioned. It depends on the definitions and possibly previously proven statements.

A condition $A$ is necessary for $B$ to be true means that $B\Longrightarrow A.$
A conditon $A$ is sufficient for $B$ to be true means that $A\Longrightarrow B.$

Despite knowing $A$ and $B$ we also need to know what $\Longrightarrow$ is, i.e. which rules of deduction are allowed. Here we have only one direction: Given $a=m/n$ prove the existence of $b$ such that $ab=ba=1.$ Proof: $b=n/m.$

Another proof would be: The rational numbers are the quotient field of the integers.

See, it all depends on what you consider allowed to be used.

 

[PeroK]

Homework Statement:: Let $a=\frac{m}{n}$ be a rational number expressed as a quotient of integers $m,n$ with $m\ne 0\;\text{and}\; n\ne 0$. Show that there is a rational number $b$ such that $\boldsymbol{ab=ba=1}$
Relevant Equations:: None, as far as I can see.

Problem statement : I cope and paste the problem as it appears in the text below.

View attachment 314974

Attempt : Not being a math student, I try and prove the above statement using an "intuitive" way.

Let us have a rational number $b = \frac{n}{m}$. Multiplying with $a$ from the right, we see $ab = \frac{m}{n}\frac{n}{m} = 1$. Since it is given that $m,n\ne 0$, we can see that a $b=\frac{n}{m}$ will always exist.

Likewise, we can multiply $a$ from the left by $b$ and show that $b$ exists.

This proves the theorem.

Issue : Am I correct with the reasoning above? A hint or correction would be welcome.

I'd say that's not a bad attempt. Yes, you explicitly let $b = \frac n m$. The thing you missed was to justify that $b$ is rational. It's obvious, of course. But, the point of these questions is to get you to use the definitions of things and basic axioms, rather than just assume things are obvious.

What two criteria determine that $b = \frac n m$ is rational?

 

[Mark44]

Since it is given that $m,n\ne 0$, we can see that a $b=\frac{n}{m}$ will always exist.

The last equation threw me off at first, until I saw the LaTeX. It looked to me like you were saying that "$ab = \frac{n}{m}$", a contradiction of what you were trying to show (i.e., that ab = ba = 1).
For clarity, you might say " a number $b=\frac{n}{m}$ will always exist.

 

[brotherbobby]

I'd say that's not a bad attempt. Yes, you explicitly let $b = \frac n m$. The thing you missed was to justify that $b$ is rational. It's obvious, of course. But, the point of these questions is to get you to use the definitions of things and basic axioms, rather than just assume things are obvious.

What two criteria determine that $b = \frac n m$ is rational?

(Sorry for the delay in reply.)

For a number $b=\dfrac{n}{m}$ to be rational, both $m$ and $n$ should not be zero ($m,n\ne 0)$. As this is already given in the problem to be the case, it shows that the number $b$ will exist.

This is additional to the small constraint on $m$ and $n$ separately : namely that $m\in \mathbb{Z}$ while $n\in \mathbb{Z_+}$. I suppose the purpose of this point is to ensure that the negative sign ("$-$") shall be taken up by the numerator.

 

[brotherbobby]

The last equation threw me off at first, until I saw the LaTeX. It looked to me like you were saying that "$ab = \frac{n}{m}$", a contradiction of what you were trying to show (i.e., that ab = ba = 1).
For clarity, you might say " a number $b=\frac{n}{m}$ will always exist.

Yes, a bad error on my part. It can read the way you said : $ab=\frac{m}{n}$.

 

[PeroK]

(Sorry for the delay in reply.)

For a number $b=\dfrac{n}{m}$ to be rational, both $m$ and $n$ should not be zero ($m,n\ne 0)$. As this is already given in the problem to be the case, it shows that the number $b$ will exist.

Not really. Both $m$ and $n$ must be integers and $m \ne 0$.

That is given in the problem, but it's still worth emphasising why you know $\frac n m$ is rational.