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What I mean to say is as follows:

Measuring the area of a surface is determining its ratio to a chosen surface called the unit of area and the chosen unit of area is a square whose side is a unit of length and if the unit of length be a metre, the unit of area will be called a square metre and similarly if the unit of length be a centimetre, the unit of area will be a square centimetre.

Then why are we using these symbols like $m^2$ or $cm^2$ to represent the unit of area when they have nothing to do with the whole procedure of measuring the areas? What’s the reason behind such a representation?

NOTE - I have asked many people the same thing and some of them gave me REASON 1 while others gave me REASON 2 but none of the reasons sounded to me accurate and I’ve explained why is it so.

REASON 1 :
They said, “Area is measured by multiplying the length and the breadth, since both are measured in terms of the unit of length therefore by multiplying the units too we end up with $m^2$ as the units of the area.”

Sounds Inaccurate Because :
This can't be a reason behind such a representation, since area is not what we get by multiplying the length and the breadth (this is rather an analogue or to be more precise it’s something that we infer from the actual procedure of measuring the areas and that too is wrongly said as it's not the product of length and breadth, it's rather the product of their numerical values ONLY).

REASON 2 :
They said, “$m^2$ is just a shorthand or an easy way to write square metre.”

Sounds Inaccurate Because :
Now this reason has two problems. Firstly, if it’s really just a shorthand then why do we chose SO SPECIFIC one and not choosing sq.m. as a shorthand (which sounds more logical and is more shorthand-oriented)?

Secondly, in the context of areas, both square metre and $m^2$ represent totally different mathematical ideas (though they sound somewhat similar while pronouncing). Square metre represents the defined unit of area i.e. a square with side length equal to 1 metre WHEREAS $m^2$ represents an arithmetical operation wherein length of $1$ metre has been multiplied with another $1$ metre length (which has nothing to do with calculation of areas).

PLEASE explain then what’s the accurate reason behind such a representation?

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If I had to pick, I'd go with your reason 1, but both have truth to them.

Yes, area is exactly length times breadth. At least for a rectangle. And most other shapes can be imagined as being decomposed into a (possibly infinite) number of rectangles, and for each of them the same rule holds.

To counter your counter-argument: Suppose you have a rectangle of $2\mathrm m\times 3\mathrm m$. You could write this as

$$2\mathrm m\times 3\mathrm m=2\times\mathrm m\times 3\times\mathrm m=2\times 3\times\mathrm m\times\mathrm m=6\times\mathrm m^2=6\mathrm m^2$$

So you really include the dimensions in that multiplication, and get back some number times the square of a length in the end. You could even feed in lengths measured in different units, and then either do a unit conversion along the way, or end up with the somewhat atypical area measurement of e.g. $\mathrm m\times\mathrm{cm}$. So it's not just the product of the numerical values, you really multiply lengths, including the units.

Using mathematical notation to combine physical units like this is common in sciences. Some examples:

  • The amount of charge provided by a battery is currency, measured in Ampere, multiplied by the time you can draw that currency, measured in hours. So the typical unit of energy is the Ampere-hour denoted as $\mathrm A\,\mathrm h$.
  • Speed can be measured in kilometers per hour. You divide a distance by the time it took you to travel this far. And how would you write this in formulas? As $\frac{\mathrm{km}}{\mathrm h}$, sometimes in-lined to km/h, and sometimes using negative exponents to denote the denominator, as $\mathrm{km}\,\mathrm h^{-1}$. Another unit of speed would be $\frac{\mathrm m}{\mathrm s}$, meters per second.
  • Acceleration is difference in speed over a given time. So the unit has to be speed divided by time, which in turn is length divided by square of time, e.g. $\frac{\mathrm m}{\mathrm s^2}=\mathrm m/\mathrm s^2=\mathrm m\,\mathrm s^{-2}$. It's probably less common to use hours instead of seconds here, but that's more a matter of convention.
  • Take potential energy as an example. If you lift a $10\mathrm{kg}$ brick $7\mathrm m$ up, you will perform work, which will be stored as energy in the elevated position of the brick. How much energy? More the heavier it is, and the higher you lift it. Also, this depends on you doing this on Earth, since lifing anything in space in free fall takes no work at all. The relevant factor here is the gravitational acceleration which is approximately $9.81\frac{\mathrm m}{\mathrm s^2}$ for Earth. The unit of energy is therefore mass times length times acceleration, or $\frac{\mathrm{kg}\,\mathrm m^2}{\mathrm s^2}$. Writing this unit as $\mathrm J$, or Joule, is merely a matter of convenience.

Regarding your reason 2: In the United States in particular, people in my experience tend to write units as “sqft” for square feet, “mph” for miles per hour, “rpm” for revolutions per minute and so on. I assume it may make it a bit easier to remember how to pronounce these units in everyday usage. But it makes mathematical calculations far more complicated. With the notation I described above, you have the same operations for numbers and for units, you can freely mix them in the same formula, and make sure to convert in the appropriate places so you can e.g. add things. With convenience notation, you have a much harder time there. So if you want to perform any non-trivial computations, the formula notation is far more explicit and convenient. I guess people in most metric countries are just too lazy to deal with two notations, one for speaking and one for computations. I know I am.

One more thought: If you add lengths, you compute things like

$$2\mathrm m+3\mathrm m=2\times\mathrm m+3\times\mathrm m=(2+3)\times\mathrm m=5\mathrm m$$

thanks to the distributive law, and if you take $3$ copies of a $2\mathrm m$ length you get

$$3\times2\mathrm m=3\times(2\times\mathrm m)=(3\times 2)\times\mathrm m=6\mathrm m$$

thanks to the associative law. So if you combine two lengths, it stays a length if you add or subtract, it becomes an area if you multiply, and for the sake of completeness it becomes a dimensionless ratio if you divide. Conversely, if you multiply a length and want the result to still be a length, you need to multiply it by a dimensionless number.

I hope my examples could shed some light on why the notation using $\mathrm m^2$ makes a lot of sense, and fits in very well with the grand scheme of things.

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Once a unit of measure is chosen, multiplication of $x \gt 0$ and $y \gt 0$ is the area of a rectangle with one side of length $x$ and the other side of length $y$. Multiplication may start off as a notation for repeated addition, but in the final analysis it represents something geometrical.

But here is something else to keep in mind:

In dimensional analysis you could have the $x\text{-axis}$ measuring meters $m$ and the $y\text{-axis}$ measuring costs in dollars $\text{USD}$ and have a cost function $y = f(x)$. The derivative $f'$ would still have meters on the $x\text{-axis}$ but the $y\text{-axis}$ would be of dimension

$\tag 1 \frac{\text{USD}}{m}$

You can take the second derivative and again only the 'dimension' of the $y\text{-axis}$ changes, becoming

$\tag 2 \frac{\frac{\text{USD}}{m}}{m}$

In physics the terminology per second per second has been accepted and, although I am not an expert in this area, I suppose we have to allow

$\tag 3 \text{per meter per meter} = m^2$

The best reason for doing this is we can simplify $\text{(2)}$ to the

$\tag 4 \frac{\text{USD}}{m^2} = = \text{USD}^1\times m^{-2}$

Compare $\text{(4)}$ to the sentence

Mathematically, the dimension of the quantity Q is given by

${\displaystyle {\text{dim}}~{Q}={\mathsf {L}}^{a}{\mathsf {M}}^{b}{\mathsf {T}}^{c}{\mathsf {I}}^{d}{\mathsf {\Theta }}^{e}{\mathsf {N}}^{f}{\mathsf {J}}^{g}}$

where a, b, c, d, e, f, g are the dimensional exponents.

found in the wikipedia article Dimensional analysis.

Although you might be able to relate $\text{per meter per meter}$ to the area of a rectangle, I find it easier to just embrace the fractional notation as a wonderful invention of the human mind.

Now to address some of the OP's concerns, if you want to attach meters as a dimension to your numbers, and you want to 'forget' that multiplication is representing the area of a rectangle, then you have to explain what you are trying to model with $x (m) \times y (m) = xy (m^2)$. Interesting, the OP really knows the answer; they stated

it's not the product of length and breadth, it's rather the product of their numerical values ONLY

So they are saying that we should not be carrying any meter dimension along in the product - numerical values are dimensionless.

However, it is perfectly legitimate to imagine a scaling operators (another name for multiplication) on the $x\text{-axis}$ measuring meters. We could scale by a factor of, say $\frac{1}{2}$ (a contraction), so that

$\tag 5 \frac{1}{2} \times \sqrt 2 \,(m) = \frac{\sqrt 2}{2}\, (m)$

and scale by a factor of $2$ (a dilation),

$\tag 6 2 \times \sqrt 2\, (m) = 2 \sqrt 2 \, (m)$

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