New contest Gain fame and stuff
Category: Dave Van Ess
Tuesday, June 24, 2008 18:00First to find the flaw in the mathematical proof wins an PSoC Eval1 board or a signed copy of my new Lab book.
1 X = Y
2 X^2 = X * Y
3 X^2 -Y^2 = X*Y - Y^2
4 (X + Y) * (X - Y) = Y * (X - Y)
5 (X+Y) = Y
6 (Y + Y) = Y (substitution from 1)
7 2 = 1
Tuesday, June 24 at 18:25 (GMT -8:00)
This is embarrassingly easy.
Line 4 to line 5 is only valid if there is no division by 0. Hence X-Y must not equal 0. Hence X must not equal Y. But this violates step 1. Hence the proof is invalid.
John Heenan
Tuesday, June 24 at 19:12 (GMT -8:00)
hm... since x=y then in line 4 it would be (x+y)*(x-y) which is always 0.
in this case if we assign ((x+y)*(x-y))/(x-y) = y, we'd come to ((x+y)*0)/0 or 0/0 which is an invalid operation. so line 4 is very dodgy one.
and from there line 5 counts of the removal of the (x-y) part (which is always 0!)
Tuesday, June 24 at 20:24 (GMT -8:00)
Looks like on line 4, the factoring is wrong. Missing an X*Y term on the right hand side.
Wednesday, June 25 at 00:44 (GMT -8:00)
It went wrong erlier (Line3):
Since x=y, x^2 - y^2 has to be 0 and
x*y - y^2 can be substituted (x=y) by x^2 - y^2 as well.
Once having a 0=0 you can even proof the pope to be a protestant.
Markus
Wednesday, June 25 at 01:50 (GMT -8:00)
Guess i am a bit late!!
Anyway for records sake, x being equal to y( x=y -step 1), x-y cant be divided from both sides in step 5 ... Damn , i want that board.
Wednesday, June 25 at 07:11 (GMT -8:00)
Nice try! Looks to me to be case of division by zero in the step from line 4 to line 5 (dividing both sides by (X-Y) which is zero) which causes the problem. As any fule kno, 2*0 = 3*0, but 2 3 !!!
Division by zero is always a dodgy thing to do - if it's allowed, then anything can be proved.
For example...
2 = 1 (from the above).
John Knox and the pope are 2 people.
Therefore John Knox and the pope are 1 person.
Hence John Knox is the pope!
Wednesday, June 25 at 10:01 (GMT -8:00)
Elementary, dear Watson!
Step [3] is obvious because reduces to 0=0; cheating is at step [4] when "cancelling" both factors (X-Y). YOU CAN'T DIVIDE BY ZERO {:-(
Well the answer is so obvious that I doubt I am giong to win anything. Just to ask you to publish something more challenging.
Thursday, June 26 at 02:16 (GMT -8:00)
step 4 multiplies zero on both side. Thats the flaw. Hope i had seen this 2 days back, I really want that book :(
Thursday, June 26 at 02:24 (GMT -8:00)
oops step 4 is a consequence of step 3 so that means step 3 is wrong as x^2=y^2
Thursday, June 26 at 06:08 (GMT -8:00)
Just to sum it up:
I assume that the (pretended) aim was to find values for the unknown x as well as for the unknown y.
You have to have two independent (!) functions for that reason.
Independent functions can't be transformed into each other, but
x=y can easily be transformed into x^2 = y*x by multiplying each side with x.
OK, since x and y have the same value there are no (!) 2 unknown.
But, none would ever write down the second eqoutation with artificial 2 unknows ... except for "baiting the trap".
Line 2 would reduce to x^2 = x^2 or x=x.
The equality of x with itself is only a fundamental mathematical presumption.
Line 3 now reduces it's statement to 0=0 (see above)
0 as factor equals every rubish you may find:
12*0 = (13+4)*0
... and can easily be substituted later, leaving any statement you want:
12= 13+4
... ignoring the fact that the substitution of 0 is a division by 0 and not allowed.
Substitution of a function by another function which is not independant (Line 6) does not lead to any sensible statement.
In this case it's just to cover up the tricks the MATHgican used before a little more.
Markus
Friday, June 27 at 11:42 (GMT -8:00)
Good afternoon, I would like to know how to activate the license for programming with the PSOC Designer "C".
agradesco your prompt response.
Thank you ..
Jonathan Acosta Salazar
Product Code Order #: 28576