The way I show it to my students is to draw two squares with side a + b. Obviously the areas of the two squares are the same. In one square draw the lines
(0,a) (b,0)
(b,0),(a+b,a)
(a+b,a),(a,a+b)
(a,a+b),(0,a)
which cuts four triangles with side a, b, c off the big square and leaves the square with side c at the middle. Area c^2
In the other big square, draw the lines (0,a),(a+b,a) and (b,0),(b,a+b) and join the intersection point to (0,0) and (a+b,a+b). That gives the same four triangles as in the first square, and squares with side a and side b, areas a^2 and b^2;
(a+b)^2 = (a+b)^2
a^2 + b^2 + 4 x (ab)/2 = c^2 + 4 x (ab)/2
a^2 + b^2 = c^2
It's far from the only proof for it, but simple, easy to draw, and makes sense.
Originally posted by Shallow BlueFabianFnas: "So you don't believe in the pythagoras theorem..."
No.
We don't believe you know the difference between calculus and trigonometry or algebra.
That's not the same thing.
Richard
Shallow Blue: "No."
Okay, then you have much studying to catch up.
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Originally posted by talzamirA fine geometrical proof, but not using calculus.
The way I show it to my students is to draw two squares with side a + b. Obviously the areas of the two squares are the same. In one square draw the lines
(0,a) (b,0)
(b,0),(a+b,a)
(a+b,a),(a,a+b)
(a,a+b),(0,a)
which cuts four triangles with side a, b, c off the big square and leaves the square with side c at the middle. Area c^2
In the other big ...[text shortened]... + b^2 = c^2
It's far from the only proof for it, but simple, easy to draw, and makes sense.
I believe in pythagoras by the way, I'm just saying that it is very difficult to prove using calculus, which is what this taxicab question comes to.
Originally posted by iamatigerThis is what the theorem by Pythagoras is: "A mathematical theorem which states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of those of the two other sides."
A fine geometrical proof, but not using calculus.
I believe in pythagoras by the way, I'm just saying that it is very difficult to prove using calculus, which is what this taxicab question comes to.
If we are not to use geometrical interpretations in the proof, then why is the theorem formulated in a geometric way from the beginning? If we are not permitted to use geometry, then we cannot prove the theorem at all.
How would you formulate the pythagorean theorem in a non-geometric way?
Or is it a trick question from you to us?
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Originally posted by FabianFnasCAAALCUUULUUUUUSSSS!!!!!
If we are not to use geometrical interpretations in the proof,
Geometry is [flash][readthisdammit]NOT!!!![/] the same thing as calculus.
If you believe it is, it is you who is ignorant if the fundaments of mathematics, not the rest of us.
Now read this again: we all realise that the Pythagorean theorem is correct in Euclidean geometry. What we do not believe is that you have a simple, obvious proof BASED ON CAAAALCUUUULUUUUSSS!!!
Clear now? Or do we have to use a steam hammer to get the point into your head?
Richard
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Originally posted by talzamirJust to remind us here is talzamir's question above.
Ask a taxi driver how far is the distance to to four blocks east and three north is, and he'll say seven. What if he's allowed a route closer to how-the-bird flies? even if you zigzag 400 times east and 300 times north, a hundredth of a block at a time....
The problem is that we have to prove that as the taxi's zigzags get tiny, the journey length becomes 5. This is like adding up lots of infinitely tiny steps, which is the same as integration, and if we could prove that it comes to 5, we would have proved pythagoras by integration, which we know is hard. We cannot do this taxicab thing with a geometrical proof because he never moves diagonally.
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Originally posted by iamatigerIn fact, I am tending towards believing that the length of the taxicabs journey is 7, because he never moves diagonally, he just takes tinier and tinier steps north and east. He cannot suddenly switch to diagonal motion at a small enough step size.
Just to remind us here is talzamir's question above.
The problem is that we have to prove that as the taxi's zigzags get tiny, the journey length becomes 5. This is like adding up lots of infinitely tiny steps, which is the same as integration, and if we could prove that it comes to 5, we would have proved pythagoras by integration, which we know is har ...[text shortened]... annot do this taxicab thing with a geometrical proof because he [b]never moves diagonally.[/b]
Originally posted by Shallow BlueOh, did I spell it wrong or something? Sorry, I'll watch out better next time. English isn't my maternal language, you know.
CAAALCUUULUUUUUSSSS!!!!!
Geometry is [b][flash][readthisdammit]NOT!!!![/] the same thing as calculus.
If you believe it is, it is you who is ignorant if the fundaments of mathematics, not the rest of us.
Now read this again: we all realise that the Pythagorean theorem is correct in Euclidean geometry. What we do not believe is that y ...[text shortened]...
Clear now? Or do we have to use a steam hammer to get the point into your head?
Richard[/b]
I ask again: The Phytagorean theorm is formulated as a geometrical theorem. Isn't it therefore natural to use geometrical arguments to prove it? If you can formulate the theorem in terms of caaalcuuuluuuuussss (correct this time?) then it is obvious that the proof can be formulated withour geometrical references.
How do you define a hypotenuse in terms of caaalcuuuluuuuussss?
I have several issues with the idea of proving pythagoras by calculus. One is that the taxi drivers journey zigzagging seems to be very close to those functions which do not differentiate (e.g. blacmange or Kock curve).
Although we are interested in integration and can use integration to find arc lengths I am not aware of a proof of the arc length formula that does not assume pythagoras.
The main objection I have though is why would you want to use calculus to prove pythagoras when there are hundreds of far more beautiful and elegant proofs for pythagoras.
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Originally posted by deriver69Because, I think, if you can simply prove this taxicab's journey length is 5, you WILL also have a simple prooof of Pythagoras with calculuuuuuus!
I have several issues with the idea of proving pythagoras by calculus. One is that the taxi drivers journey zigzagging seems to be very close to those functions which do not differentiate (e.g. blacmange or Kock curve).
Although we are interested in integration and can use integration to find arc lengths I am not aware of a proof of the arc length formu ...[text shortened]... rove pythagoras when there are hundreds of far more beautiful and elegant proofs for pythagoras.
This is one analysis about the cab-drivers paradox.
Original problem: "Ask a taxi driver how far is the distance to to four blocks east and three north is, and he'll say seven." Is this true?"
Yes of course, if you ask him.
Let's say that instead of blocks we use the distance in units. How long will the cab drivers drive? Answer seven units. As 4+3=7.
Now, half that distances, and we get 4/2+3/2 and we get 7/2. But in order to let the cab driver reach his destination, he must do this twice. (4/2+3/2)*2 is still equal to 7 units.
If we make the same with division by 3, 4 or 5 and further we get the same result, namely 7. Even if we substitute the denominator with n and let n be whatever number, (4/n+3/n)*n we get still get 7, of purely aritmetic reasons.
Let's make a serie out of this:
s(1) = (4/1+3/1)*1 = 7
s(2) = (4/2+3/2)*2 = 7
s(3) = (4/3+3/3)*3 = 7
...
s(n) = (4/n+3/n)*n = 7
If we let n go from 1 all the way to the highest thinkable number we still get the answer 7.
And finally: If we let n -> infinity, what will happen to our seven? Does it suddenly turn to five?
Anwer: No, it doesn't.
However, if we let the cab driver drive his cab and draw a line as he drives, then we get a hypotenuse looking line which resembles a straight line when the numbers of turns (n) is going upwards high enough. But if the line is infinitely thin, and you zoom it in sufficiently much, then you will still see the eastward and northward lines.
What looks like a straight line, isn't. And that's the paradox explained.
(Btw, I've never said that n will actually reach infinity.)
