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Subject: "Proof #23 using Wolfram Alpha"     Previous Topic | Next Topic
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jmolokach
Member since Aug-17-10
 Nov-22-10, 01:00 PM (EST)
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"Proof #23 using Wolfram Alpha"
 
   I have been having fun using Wolfram Alpha. This is likely an identical manifestation of proof #23 of the Pythagorean Theorem, but thanks to Wolfram for doing the "with a little effort" part...

This comes from setting (1/2)ab = Heron's formula for a right triangle with legs a and b. "If any proof deserves to be called algebraic, this one does..."

https://www.wolframalpha.com/input/?i=solve%282ab%3Dsqrt%28%28a%2Bb%2Bc%29%28b%2Bc-a%29%28a%2Bc-b%29%28a%2Bb-c%29%29%2Cc^2%29

molokach


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  Subject     Author     Message Date     ID  
  RE: Proof #23 using Wolfram Alpha alexbadmin Nov-22-10 1
     RE: Proof #23 using Wolfram Alpha jmolokach Nov-22-10 2
         RE: Proof #23 using Wolfram Alpha alexbadmin Nov-22-10 3
             RE: Proof #23 using Wolfram Alpha jmolokach Nov-22-10 4
                 RE: Proof #23 using Wolfram Alpha alexbadmin Nov-22-10 5
                     RE: Proof #23 using Wolfram Alpha jmolokach Nov-22-10 6
                         RE: Proof #23 using Wolfram Alpha alexbadmin Nov-22-10 7
                             RE: Proof #23 using Wolfram Alpha jmolokach Nov-22-10 8
                                 RE: Proof #23 using Wolfram Alpha alexbadmin Nov-22-10 9

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alexbadmin
Charter Member
2666 posts		 Nov-22-10, 01:20 PM (EST)
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1. "RE: Proof #23 using Wolfram Alpha"
In response to message #0
 
   That's very nice. But what does c^2 do at the end of the equation?


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jmolokach
Member since Aug-17-10
 Nov-22-10, 03:26 PM (EST)
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2. "RE: Proof #23 using Wolfram Alpha"
In response to message #1
 
   I do not think the link pasted correctly.

The result given in Wolfram Alpha for c is:

c = - sqrt(a^2 + b^2)
c = sqrt(a^2 + b^2)

I have attached a picture of the results from https://wolframalpha.com

molokach

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/4cead0ce542f6b70.jpg

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alexbadmin
Charter Member
2666 posts		 Nov-22-10, 03:28 PM (EST)
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3. "RE: Proof #23 using Wolfram Alpha"
In response to message #2
 
   You asked to solve something. At the end of this expression was an extra argument c^2. What was it for?


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jmolokach
Member since Aug-17-10
 Nov-22-10, 03:45 PM (EST)
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4. "RE: Proof #23 using Wolfram Alpha"
In response to message #3
 
   Originally I asked to solve the equation for c^2. Apparently it would only give me the solution for c, so I suppose that is something attributed to this web software version of Mathematica... You must only be able to solve for a single variable. But nonetheless it gives the desired result.

molokach


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alexbadmin
Charter Member
2666 posts		 Nov-22-10, 03:47 PM (EST)
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5. "RE: Proof #23 using Wolfram Alpha"
In response to message #4
 
   I see.

It's really a very clever piece of software. It figures out what to solve for from the form of the equation.

a and b appear on both sides while c only on one. So it'solves for c even if there is no c^2.

Curious.


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jmolokach
Member since Aug-17-10
 Nov-22-10, 04:36 PM (EST)
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6. "RE: Proof #23 using Wolfram Alpha"
In response to message #5
 
   You should try it.

Go to https://wolframalpha.com

Type in the search box:

Solve 2ab=sqrt((a+b+c)(b+c-a)(a+c-b)(a+b-c)) for c

Watch what it does...

In some cases (not ones as complex as this though), this thing will actually show the steps to solving an equation!

See this link for more on that:

https://blog.wolframalpha.com/2009/12/01/step-by-step-math/

molokach


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alexbadmin
Charter Member
2666 posts		 Nov-22-10, 04:37 PM (EST)
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7. "RE: Proof #23 using Wolfram Alpha"
In response to message #6
 
   >You should try it.
>
Well, how do you think I knew that c^2 was not necesary?


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jmolokach
Member since Aug-17-10
 Nov-22-10, 07:33 PM (EST)
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8. "RE: Proof #23 using Wolfram Alpha"
In response to message #7
 
   Sorry I thought you noticed it in the url I posted...

molokach


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alexbadmin
Charter Member
2666 posts		 Nov-22-10, 07:35 PM (EST)
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9. "RE: Proof #23 using Wolfram Alpha"
In response to message #8
 
   >Sorry I thought you noticed it in the url I posted...

I do not understand. This is when I noticed it. The url was mangled, but it was not difficult to fix it.

As I said, it's a wonderful software and very clever at that.


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