Invalid proofs of geometry

Why is this not a valid proof of the pythagorean theorem.

#Geometry #ProofTechniques #Proofs

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Can it be generalized for all possible pairs of squares?

Daniel Liu - 7 years, 3 months ago

I give up. Can you explain why it is not a valid proof?

Silas Hundt Staff - 7 years, 3 months ago

Yeah, this is how I was taught in school actually (well, everyone already knew it, but the teacher thought it was a cool thing to help explain why it works).

Sam Solomon Staff - 7 years, 3 months ago

I am not sure, when I searched online I found this as the reason. While the proof looks like many other proofs by dissection and rearrangement, this one is not complete in that the construction does not go through where one of the legs of the given triangle is essentially less than the other This is the website:http://www.cut-the-knot.org/pythagoras/FalseProofs.shtml refer to proof #3

Mardokay Mosazghi - 7 years, 3 months ago

The main concern is that the picture doesn't deal with all possible cases. In this picture, we see that the small square is broken up into 3 vertical strips, and we can see how to rearrange them. The central strip juts nicely into the brown and blue triangles, with the green triangles placed below.

However, there is no reason why there are only 3 vertical strips. As the larger base grows larger, the smaller square could take many many vertical strips, after which it is unclear how to glue everything together. What is the role of the brown and blue triangles, and where do they go?

This is akin to drawing a special picture for a geometry question, say by setting the triangle to be equilateral or isosceles, and finding the answer for the specific case. We know that the answer is only valid in that scenario, and not in the more general case.

Calvin Lin Staff - 7 years, 3 months ago

Thanks

Mardokay Mosazghi - 7 years, 3 months ago

solve the theorem in the middle of square 3/5^2+4/5^2=25 make sq of 25=9+16=25

amar nath - 7 years, 1 month ago

TRANGLE IN WHICH SIN=4/5^2=16/25 AND COS=3/5^2=9/25 SO EQUAL TO 25/25=1 THEREFORE 25=25 AFTER WHICH LEFT DISOLVE INTO 4^2+3^2=25 .IF ASSUME SIDE OF MIDDLE TRIANGLE ARE 3,4.,HYPOTINEOUS=5. PLEASE COMMENT.

amar nath - 7 years, 1 month ago

all the sq are divided into right angle triangle. if the sq are of side are 3,4,5 , it will comes equal to the area of sq root 50 after measurement of each triangle . central triangle. figure cannot be drawn on same sheet ihave roof area equal to the three sq.

amar nath - 7 years ago

my answer was 0.2 pleace comment differnce 0.217& 0.2

amar nath - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$