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Why is n (a number) = $\frac{n}{1}$?

My reasoning is as follows:

For a fraction $\frac{a}{b}$, b (denominator) is the number of equal parts the whole (1) is divided into and a (numerator) counts how many of these parts we're interested in. So e.g. $\frac{2}{5}$, read as two-fifths, represents 1 broken up into 5 equal parts and taking 2 of them.

Quite naturally then n = $\frac{n}{1}$ as $1$ is "divided" into 1 equal "parts" which is again 1 and we take n of them (n $\times$ 1 = n).

Ergo, my brain tells me n = $\frac{n}{1}$

Is my argument sound? Is there a better (simpler/shorter/correct) proof?

Muchas gracias.

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    $\begingroup$ Your way of thinking is basically correct. In more "mathematical" terms, if $b \not = 0$, the fraction $\frac{a}{b}$ is the unique number characterized by the property that if you multiply it by $b$, you recover $a$. In other words, $\frac{a}{b}$ is the unique solution of the equation $bx = a$ with unknown $x$. In particular, $\frac{n}{1}$ is the only solution of the equation $1\times x = n$. Since $n$ is clearly also a solution of this equation, by unicity we have $\frac{n}{1} = n$. $\endgroup$
    – Suzet
    Commented Dec 17, 2022 at 2:50
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    $\begingroup$ I'm not really sure why this got multiple downvotes. It's an elementary question, but there's nothing wrong with elementary questions. $\endgroup$
    – Noah Schweber
    Commented Dec 17, 2022 at 3:11
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    $\begingroup$ The number $1$ is the identity element so $1\cdot a=a$ for all $a$. An inverse for a number $a$ is a number called $a^{-1}$ with the property that $aa^{-1}=a^{-1}a=1$. We also write $a^{-1}$ as $\frac{1}{a}$. Now, $1$ is its own inverse so $1=\frac{1}{1}$ and $1\cdot n=n$. Therefore $n=\frac{1}{1}\cdot n=\frac{n}{1}$. $\endgroup$
    – John Douma
    Commented Dec 17, 2022 at 4:24
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    $\begingroup$ Suzet's proof is just much better than either your argument in the post, or re-interpretation in the comments. It comes directly from the definition of division and the definition of $1$. $1$ is the multiplicative identity: $1\times x = x$ for any $x$. And $\frac ab$ is defined as the number such that $b\times \frac ab = a$. All your concepts are much further removed from these definitions, require additional facts to be proven, and are more complicated. $\endgroup$
    – Paul Sinclair
    Commented Dec 18, 2022 at 4:21
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    $\begingroup$ Yes. That is what I am referring to. The first is the definition of division: $\frac n1$ is the number which when multiplied by $1$ gives $n$, and the second is the definition of the multiplicative identity $1$: multiplying $n$ by $1$ gives $n$. $\endgroup$
    – Paul Sinclair
    Commented Dec 18, 2022 at 6:23

2 Answers 2

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A simpler and more straightforward proof flows from the multiplicative identity axiom and the definition of divides.

For any real numbers $a$ and $b$, we say $a$ divides $b$ and write $\frac{b}{a} = k$ if and only if there exists a real number $k$ such that $ak = b$. In other words, you may interpret the statement $\frac{b}{a} = k$ as saying exactly $k$ multiplies of $a$ equal $b$.

According to the multiplicative identity law, we have $n \cdot 1 = n$ for every real number $n$. So, by definition of divides, $1$ divides $n$ and we may write

$$ \frac{n}{1} = n $$

meaning exactly $n$ multiples of $1$ equal $n$. The multiplicative identity law also implies $n$ divides $n$ and we may write

$$ \frac{n}{n} = 1 $$

meaning exactly $1$ multiple of $n$ equals $n$.

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  • $\begingroup$ Arigato, but the assumption here is that $\frac{n}{1} = n \div 1$. That isn't "permitted" in the system I'm in right now. You see if $a \div b = \frac{a}{1} \div \frac{b}{1}$ then (note here that $a = \frac{a}{1}$ has to be an independent theorem), then I can do this: $\frac{a}{1} \times \frac{1}{b} = \frac{a}{b}$, proving $a \div b = \frac{a}{b}$. I want to avoid a circulos in probando. $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 5:50
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    $\begingroup$ @AgentSmith What system are you in? It doesn't seem to be mentioned in the question. If you want a non-circular proof then you need to give more precise definitions of the notations and terms you are using; also a decent listing of the facts that may be used in the proof. $\endgroup$
    – David K
    Commented Dec 31, 2023 at 7:07
  • $\begingroup$ Apologies for the delay. The mathematical equivalent of baby logic i.e. baby math. @DavidK. A very soulcrushing name for those who have to struggle. $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 11:10
  • $\begingroup$ @AgentSmith Could you provide a link to where this system of baby logic is described? I haven't heard of it before, and I couldn't find it after performing an internet search. The derivation of $\frac{n}{1} = n$ above proceeds directly from the definition of divides and an axiom of the real numbers known as the multiplicative identity law, so there is no circularity here. In other words, I am not assuming the very thing I'm trying to prove. You can go on to prove something like $\frac{b}{a} = \frac{b}{1} \cdot \frac{1}{a}$, and it all rests on the definition and axiom I just mentioned. $\endgroup$
    – RyRy the Fly Guy
    Commented Dec 31, 2023 at 14:45
  • $\begingroup$ We may not use $\frac{a}{b} = k$ because it assumes $a \div b = \frac{a}{b}$ (this is what I want to prove). By the way, spasibo. $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 19:27
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To answer such question you need to define what $\frac{a}{b}$ symbol means. Math is a science of precision.

So in a typical setup you would have a pair $(R,\cdot)$, where $R$ is a set and $\cdot$ is a function $R\times R\to R$ typically called "multiplication". Of course "set" is a primitive object, while function and Cartesian product have their own definitions as well.

Additionally we would have a special element $1\in R$ such that $x\cdot 1=1\cdot x=x$ for any $x\in R$. Then you can prove that such $1$ is unique if it exists. This element is often called neutral or multiplicative identity.

Finally the symbol $\frac{a}{b}$ for $a,b\in R$ is defined as $a\cdot b^{-1}$, where $b^{-1}$ is defined as an element of $R$ such that $b\cdot b^{-1}=b^{-1}\cdot b=1$, i.e. the inverse of $b$. This is under assumption that $\cdot$ is commutative (not a strict requirement but without it $\frac{a}{b}$ behaves in a weird, counterintuitive way, e.g. $\frac{a}{b}\cdot\frac{c}{d}$ is not necessarily equal to $\frac{a\cdot c}{b\cdot d}$). Again if $b$ has an inverse then it has to be unique.

Note that the symbol "$-1$" here is not related to the special neutral element $1\in R$. These are two different things that share the same symbol. Also "$x^{-1}$" does not refer to exponentiation, it literally stands for "the inverse of $x$". Both might be confusing but it is a standard mathematical convention to use such symbols. Those two are connected with the standard meaning, but that's a story for another time.

With all of that we easily see that $1\cdot 1=1$ and therefore $1^{-1}=1$. And so for any $n\in R$ we have

$$\frac{n}{1}=n\cdot 1^{-1}=n\cdot 1=n$$

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  • $\begingroup$ Nice! Circularity-meter is giving a reading. Perhaps mine is broken. Not a problem though, it's easy to lose your bearings in what is a jungle of ideas. $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 11:19
  • $\begingroup$ @AgentSmith I assure you there is no circularity involved here. Feel free to ask questions if anything is unclear. $\endgroup$
    – freakish
    Commented Dec 31, 2023 at 12:42
  • $\begingroup$ It's a roundabout way of using $a \times \frac{1}{a} = 1$. A necessary requirement would be an alternative and yet valid route (consistent with how fractions behave). About the circularity ... you're using exponents, and that comes after fractions. Top-down, ok, but bottom-up better, oui? $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 19:21
  • $\begingroup$ @AgentSmith I am not using exponents. You misinterpreted what "$x^{-1}$" actually is, even though I precisely defined it in the answer. This has nothing to do with exponentiation. Perhaps I wasn't clear enough, I've updated the answer. $\endgroup$
    – freakish
    Commented Dec 31, 2023 at 19:47
  • $\begingroup$ I see. So we have the identity element for multiplication, $1$ and what you're employing here is the multiplicative inverse, for a number $n$, it would be $n^{-1}$ such that $n \times n^{-1} = 1$. That would mean $n \div 1 = n \times 1^{-1} = n \times 1 = 1$. What about $\frac{n}{1}$? Before I use your definition, I have to be shown $\frac{n}{1} = n \div 1$. $\endgroup$
    – Hudjefa
    Commented Dec 31, 2023 at 19:53

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