Difference between revisions of "Nichomauss' Theorem"
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==Nichomauss' Theorem== | ==Nichomauss' Theorem== | ||
| − | Nichomauss' Theorem states that <math>n^3</math> can be written as the sum of <math>n</math> consecutive integers, thus giving us <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>. | + | Nichomauss' Theorem states that <math>n^3</math> can be written as the sum of <math>n</math> consecutive integers, thus giving us <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>.WE ARE CRAZY |
==A Visual Proof== | ==A Visual Proof== | ||
Imagine a cuboid with a height of <math>1</math>, and length and width of <math>n</math>. Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have <math>l=w=h=1</math> (call this <math>a_1</math>). Now, extend its length and width by <math>2</math> to get another cuboid with <math>l=w=3</math> and <math>h=1</math> (call this <math>a_2</math>). Color <math>a_1</math> yellow and <math>a_2</math> blue. Now, it's easy to see that <math>a_2</math> has <math>(3 \cdot 3 \cdot 1)-(1 \cdot 1 \cdot 1)=8</math> unit cubes and <math>a_1</math> has <math>1 \cdot 1 \cdot 1=1</math> unit cube. We can then rearrange the <math>8</math> unit cubes into a <math>2 \cdot 2 \cdot 2</math> cube. Thus, we clearly have <math>(1+2)^2=1^3+2^3</math>. And we can continue this process to <math>n</math> unit cubes and <math>n</math> cuboids with <math>l=w=n</math>, <math>h=1</math>, which gets us to <math>(1+2+...+n)^2=1^3+2^3+...+n^3</math>. | Imagine a cuboid with a height of <math>1</math>, and length and width of <math>n</math>. Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have <math>l=w=h=1</math> (call this <math>a_1</math>). Now, extend its length and width by <math>2</math> to get another cuboid with <math>l=w=3</math> and <math>h=1</math> (call this <math>a_2</math>). Color <math>a_1</math> yellow and <math>a_2</math> blue. Now, it's easy to see that <math>a_2</math> has <math>(3 \cdot 3 \cdot 1)-(1 \cdot 1 \cdot 1)=8</math> unit cubes and <math>a_1</math> has <math>1 \cdot 1 \cdot 1=1</math> unit cube. We can then rearrange the <math>8</math> unit cubes into a <math>2 \cdot 2 \cdot 2</math> cube. Thus, we clearly have <math>(1+2)^2=1^3+2^3</math>. And we can continue this process to <math>n</math> unit cubes and <math>n</math> cuboids with <math>l=w=n</math>, <math>h=1</math>, which gets us to <math>(1+2+...+n)^2=1^3+2^3+...+n^3</math>. | ||
Revision as of 20:20, 17 November 2020
Nichomauss' Theorem
Nichomauss' Theorem states that 
can be written as the sum of 
consecutive integers, thus giving us 
.WE ARE CRAZY
A Visual Proof
Imagine a cuboid with a height of 
, and length and width of . Divide it into unit cubes. Now, starting from the bottom right of the cuboid (when it's flat on the ground), imagine the first cube to have 
(call this 
). Now, extend its length and width by 
to get another cuboid with 
and 
(call this 
). Color yellow and blue. Now, it's easy to see that has 
unit cubes and has 
unit cube. We can then rearrange the 
unit cubes into a 
cube. Thus, we clearly have 
. And we can continue this process to unit cubes and cuboids with 
, , which gets us to 
.
