I just proved the pythagorean theorem

[catatomic]

And I checked hundreds of them and it looks original.

 

[White 6]

OK. Clean it up (if necessary) and post a pic of your proof, to be judged by the math minded members.
The theorem has been proved numerous times by many different methods – possibly the most for any mathematical theorem. Wikipedia

 

[OKTexas]

And I checked hundreds of them and it looks original.

What looks original?

.

 

[catatomic]

Unfortunately, my proof does not appear to work at this time.

 

[catatomic]

It's embarassing - I haven't been to grad school in almost 20 years, but I will share my proof anyhow since someone else might be able to make it work:

In fig. 1, we have a square for a, then increase the size of the square by an area of b2, the sum of areas is a2+b2. The extra part is ax, x2 and ax.

In fig. 1, we have a square for b, then increase the size of the square by an area of a2, the sum of areas is a2+b2. The extra part is by, y2 and by.

If we look at a,b and c in fig.s 3 and 4, it looks like c=a+x or c=b+y. If we could prove that then we could prove the Pythagorean theorem.

Notice that (a+x)2 = a2 + b2 = (b+y)2. (1) They were all the same thing. If c=a+x and c=b+y, 2C2= (a+x)2 + (b+y)2 = a2 + 2ax + x2 + b2 + 2by + y2. (2) But these are all the parts in fig. 1 and fig. 2, which each add to a2+b2, so 2c2 = 2a2+2b2 or c2 = a2+b2. (3) See fig. 5.

Now we need to gather that c=a+x. c ≤ a+x by area because c must be in the square (fig. 1/fig. 2) because a and b are in the square. That gives a maximum area of the diagonals. On the sides 1x1=1. On the diagonals √2*√2/2 = 1. Thus c ≤ a+x. (4)

Now we need to gather that c ≥ a+x. Or c ≥ b+y (both are equivalent).

From figs 1 and 2, b2 = 2ax+x2, and a2 = 2by+y2. (5)

Adding, a2+b2 = 2ax + 2by + x2 + y2. (6)

C2 =? 2ax + 2by + x2 + y2 (7)

C2 ≥? a2 + 2ax + x2 (8)

C2 ≥? b2 + 2by + y2 (9)

This leads to 2ax+2by+x2 +y2 ≥? a2 + 2ax + x2 (10)

And 2by + y2 ≥? a2. (11)

We also get 2ax+2by+x2 +y2 ≥? b2 + 2by + y2 (12)

And 2ax + x2 ≥? b2. (13)

We solve quadratic equations

Y= (-2b + √(4b2+8a2b))/2 (14)

X = (-2a + √(4a2+8ab2))/2 (15)

The roots must be positive. Hence 4b2+8a2b > 0 and 4a2+8ab2 > 0. (16,17)

4b2 > -8a2b (18)

This is guaranteed when b>0.

Y will be > 0 because √4b2 already makes the -2b positive.

4a2 > - 8ab2 (19)

This is guaranteed when a>0.

X will be >0 because √4a2 already makes the -2a positive.

By the quadratic formulas (14) and (15), x = greater than -a and y = greater than -b, so if c>0, c ≥ a+x and c ≥ b+y and at the boundary point c=a+x =b+y. (20).

Note that c,a and b work >0 and then x and y will be.

(4) and (20) mean that c=a+x and then (1),(2) and (3) mean that C2=a2+b2

Then by (3) we solved the problem.

We could have more easily said that C2 = (a+x)2 = a2+b2.

 

[catatomic]

Try this:

Thread 'Proof of the Pythagorean Theorem another go'

Jul 13, 2025

Previous attempt I just proved the pythagorean theorem

In fig. 1, x is defined so that ax+ax+x2 = b2, and in fig. 2 y is defined so that by+by+y2=a2.

Fig. 1 and Fig. 2 have the same area a2+b2, and they are square, so they have the same edges in each and together. a2+02=c2 and 02+02=02 are trivial. The edges are all a+x or b+y.

Fig. 3 and 4 suggest c=a+x and c=b+y.

c will fit in the square because a and b fit in the square (fig. 5). WLOG (without loss of...

• catatomic
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• Forum: General Discussion