Root 2 irrationality proof (geometric)

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SUMMARY

The forum discussion centers on the geometric proof of the irrationality of the square root of 2, specifically addressing confusion around certain terminologies and steps in the proof. Key points include the clarification of "swinging a b-leg to the hypotenuse," which refers to drawing an arc of a circle with radius b from point R to the hypotenuse. Additionally, the expression \(\dfrac{2b-a}{a-b}\) is analyzed, demonstrating that both \(2b-a\) and \(a-b\) are indeed smaller than a and b, respectively, thus leading to a contradiction in the proof.

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phospho
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I was looking over this proof and I have some questions:

http://jeremykun.com/2011/08/14/the-square-root-of-2-is-irrational-geometric-proof/

Second paragraph, what does "swinging a b-leg to the hypotunese" mean? Also, where did the arc come from, I really don't understand

also, the last part ## \dfrac{2b-a}{a-b} ## how does this lead to a contradiction (i.e. how is 2b-a and a-b smaller than a and b?

really don't understand the proof overall, if someone could walk me through it or give me some pointers it'd really help thank you
 
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hi phospho! :smile:
phospho said:
Second paragraph, what does "swinging a b-leg to the hypotunese" mean? Also, where did the arc come from, I really don't understand

it's a really stupid way of saying "draw an arc of a circle of radius b from R to the hypotenuse" :redface:
i.e. how is 2b-a and a-b smaller than a and b?

2b - a = b + (b -a), which is obviously < b :wink:

(and a - b < a)
 

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