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This might sound really stupid, but is there a proof for the volume of a cube, i.e $V = x^3$, the only way I can think it through in my head is that there is going to be $x^3$ unit cubes in a cube of length $x$ meaning the volume is $x^3 * 1$, however I'm not sure what the actual proof is. (I can't find any online)

Nav Bhatthal
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3 Answers3

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A proof depends on what you are allowed to assume.

For example, let's assume we know that if you start with a cuboid and extend it by a factor $x$ in one direction only, you multiply the volume by a factor of x.

Then take a unit cube (of volume assumed to be $1$) oriented to standard axes (choose the axes to fit the cube) and extend it by a factor $x$ in the $X$ direction, then take the resulting cuboid and extend it by $x$ in the $Y$ direction, and then by $x$ in the $Z$ direction, then each side is of length $x$ and the volume is $x^3$.

This is a bit in the spirit of Euclid, who builds up a theory of area in this kind of way.

You can go back another stage and prove that extending by a factor of $x$ in one direction increases the volume by a factor of $x$ using congruence and making the assumption that congruent shapes have the same volume. Then you do it first for $x$ integral, then for $x$ rational (so to get 5/3 you extend by 5 and divide the result into three congruent boxes) and then you do Real values of $x$ by taking limits (and you need the theory of the Reals to do that).

Congruence was the base of the theory of measure before integration - integration extends congruence as originally understood. There are some subtleties involved in doing so - see, for example this on the Banach Tarski Paradox for counterintuitive congruences.

Mark Bennet
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  • It is also easier to use congruence in two dimensions than in three - for which Dehn proved (Hilbert's third problem) that two polyhedra of the same volume need not be dissectible into a collection of congruent pieces. This meant that a new idea had to be introduced to extend the notion of measure from two dimensions to three. See this Numberphile video for more information. https://www.youtube.com/watch?v=eYfpSAxGakI – Mark Bennet Nov 23 '22 at 21:37
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If you have a grid and you want to count the number of rectangles in that grid, you multiply the number of columns by the number of rows.

Similarly, when you want to calculate the area of a rectangle, you count the number of square centimeters in that area by multiplying the number of columns by the number of rows, which happen to be the lengths of corresponding sides measured in centimeters. If you can't visualize this try drawing it.

Now taking that to the third dimension, you find the number of cubic centimeters in a cuboid by multiplying the number of columns, rows, and layers in the cuboid, and those correspond to the cuboid's length, width and height.

Anas Khaled
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  • So it's just a matter of finding how many unit cubes fit into said cube? – Nav Bhatthal Nov 24 '22 at 05:38
  • Yes, that's how I think about it, other ideas were purposed here that are indeed interesting, which is why I'm glad you asked this question. – Anas Khaled Nov 24 '22 at 09:40
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    It's something that I had literally never bother to think about but once I did I wouldn't stop untill I got an ok answer. Never stop asking. – Nav Bhatthal Nov 24 '22 at 11:36
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Consider a cube having base area $A$ and height $x$ . Move some $h$ distance upward ( $0<h<x$) and consider a small cuboid having height d$h$. Essentially, volume of this cuboid is $A$d$h$. Now summing up these small volumes , we have ;

$ V = \int_0^x Adh= Ax$

Similarly, the value of $A$ can be computed to be $x^2$ and hence the volume.

CAUTION: It is a wrong intuition of thinking $Ax$ as "we are multiplying area by $x$ because height is $x$ and hence adding $A$ for $x$ times gives volume "- because here $x$ do not denotes discrete entities. However, it is valid for the elemental volume $A$d$h$.

An_Elephant
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