## Most recent change of Fallacy

Edit made on April 20, 2009 by derekcouzens at 07:36:33

Deleted text in red / Inserted text in green

WW
A Fallacy in Mathematics is an apparently correct logical argument which leads to an incorrect conclusion.

COLUMN_START^
[[[
!! Fallacy 1: 1+1=1

Let EQN:a=b
* EQN:a^2=ab
* EQN:a^2-b^2=ab-b^2
* EQN:(a+b)(a-b)=b(a-b)
* EQN:a+b=b
* EQN:a+a=a (as a=b)

Now let a=1, and hence 1+1=1.
]]]
COLUMN_SPLIT^
[[[
!! Fallacy 2: -1 = 1

* EQN:sqrt{-1}=sqrt{-1}
* EQN:sqrt{\frac{-1}{1}}=sqrt{\frac{1}{-1}}
* EQN:\frac{sqrt{-1}}{sqrt{1}}=\frac{sqrt{1}}{sqrt{-1}}
* EQN:sqrt{-1}sqrt{-1}=sqrt{1}{sqrt{1}
* Hence EQN:-1=1
]]]
COLUMN_END
[[[> IMG:angles1.png ]]]
!! Fallacy 3: all angles are right angles
(Rouse Ball's Fallacy)

Construct a quadrilateral ABCD such that AC = BD and angle CAB is a right angle and angle DBA is obtuse.

* AB is therefore not parallel to CD,
** hence perpendicular bisectors are not parallel
** hence perpendicular bisectors intersect at some point: call it E

* Construct the perpendicular bisector of AB namely ME
* Construct the perpendicular bisector of CD namely NE

* Triangle AEM is congruent to triangle BEM (RHS)
** thus angle EAM = angle EBM

* Triangle ACE is congruent to triangle BDE (SSS)
** thus angle EAC = angle EBD

* Subtracting these angles gives angle CAB = angle DBA

Therefore all obtuse angles are right angles.

Similarly all acute angles are right angles (complement of obtuse angles) ... all angles are right angles.