@plasmid, With the givens, I do not believe {2, 2Π, 4 - Π} satisfy the three reals.

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# real number and blackboard

### #21

Posted 26 November 2014 - 04:38 PM

### #22

Posted 26 November 2014 - 10:54 PM

(4)*(2) + (-1)*(2pi) + (-2)*(4-pi) = (8) + (-2pi) + (-8 + 2pi) = 0

Solving for pi...

a(2) + b(2pi) + c(4-pi) = 0

2a + 2b*pi + 4c - c*pi = 0

2a + 4c + (2b-c)pi = 0

implies 2b-c = 0, which it does

### #23

Posted 27 November 2014 - 04:53 AM

@plasmid, With the givens, I do not believe {2, 2Π, 4 - Π} satisfy the three reals.

Spoiler for

This is an interesting puzzle..

### #24

Posted 27 November 2014 - 06:13 AM

Yes, division by zero does create a situation where all values are correct (and thus undefined). In addition to this error, I see where I had an error while calculating.

**Edited by DejMar, 27 November 2014 - 06:18 AM.**

### #25

Posted 27 November 2014 - 07:02 PM

### #26

Posted 28 November 2014 - 10:44 AM

Your three reals {2, 2pi, 4-pi} did show that the proposition that all terms will be a multiple of the irrational value if an irrational value did exists is not necessarily true. (Hence a proposition and not an assertion). What you did demonstrate is that not only scaling, the multiplicative aspect, is an operational factor to be considered, but that the additive aspect employed is a factor, as well.

**Edited by DejMar, 28 November 2014 - 10:45 AM.**

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