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When formalizing the notion of area, among other postulates the area of a unit square can be defined to be $1$. Using this we can prove that the area of square of length $x$ has area $x^2$. Here a link for such a proof.

What if we define it to be something other than 1? Say 2, the formula for area of a square will then becomes $2x^2$. Using which we can derive the formulas for areas of other shapes.

Can something go wrong? Is the choice of $1$ as the multiplicative factor arbitrary?

Note : By Unit Square I mean a square of side length 1.

mihir jain
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    It will be no more the "unit square" but the "pair square" – Mauro ALLEGRANZA Mar 08 '21 at 12:56
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    Yes, if you define the unit square area to be $2$, then all your areas will be double the usual areas. In fact, for some purposes, the normalization may be taken so that the unit disk has area $1$. But still, all areas in that system are a constant $1/\pi$ times the standard areas. – GEdgar Mar 08 '21 at 12:56
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    It is a good idea to be consistent. If the measure of the side of the square is $1$, but its area is $2$, and so every area (but not length) is doubled, this can work (as in principle the measures of lengths and of areas are independent to each other) - but I would say these two choices don't "go well" with each other. (The formula of the area of rectangle would be $A=2ab$, the unit of area won't be $m^2$ but a new "unit of area"...) Or, why not then redefine the length of this square to be $\sqrt{2}$ instead of $1$ - and you get back to a (smaller) unit square of area $1$? –  Mar 08 '21 at 13:33
  • Definitions are not made arbitary, but in such a way that logical structures emerge. In some cases, we have freedom, in other, like here, there is only one reasonable choice. What else should a UNIT-area be than an area with magnitude $1$ (whatever this $1$ means) ? If every mathemtician would create his/her own definitions, communications between mathematicians would become utterly impossible. – Peter Mar 08 '21 at 13:50
  • Let us transform this idea to the discrete case : The smallest positive integer is $2$ , not $1$ and we count $2,4,6,8,\cdots$. We now have only even numbers, only one prime number , therefore no prime factorizations (at least non in the usual sense) etc. – Peter Mar 08 '21 at 13:51
  • See, for instance, my answer for the Math.SE question "Area in axiomatic geometry". – Blue Mar 08 '21 at 20:03

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Without that axiom, we cannot deduce the area of anything (with nonzero area). We can prove statements like "this polygon has $3.5$ times the area of that other polygon" or "the area of this region is the difference between the areas of those other two regions", but we are limited to relative statements, not absolute ones.

As a result, it is consistent with the other axioms for any square to have area $1$. So we pick one square, call it the unit square, and decide that it will have area $1$. (We could instead pick a square and decide it will have area $2$; we probably shouldn't call it a "unit" square, though.)

Another problem this axiom solves is that it tells us that not everything has area $0$. (That's otherwise consistent with the other axioms for area, in most formulations.)

Misha Lavrov
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  • Thanks for the answer. I thought the "unit" in unit square comes the fact that it has sides of unit length. I was wrong. – mihir jain Mar 08 '21 at 13:09
  • Well, I'm not sure the "unit" in unit square comes from any single thing. Before anyone was doing the axiomatic approach very carefully, I bet people simply said "A unit square is a square with side length and area $1$". So the terminology (and everyone) would be very confused if these two properties were not on the same squares. – Misha Lavrov Mar 08 '21 at 13:36