Fallacy

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Fallacy 1: 1+1=1 Let $a=b$ $a^2=ab$ $a^2-b^2=ab-b^2$ $(a+b)(a-b)=b(a-b)$ $a+b=b$ $a+a=a$ (as a=b) Now let a=1, and hence 1+1=1.

Fallacy 2: -1 = 1 $sqrt{-1}=sqrt{-1}$ $sqrt{\frac{-1}{1}}=sqrt{\frac{1}{-1}}$ $\frac{sqrt{-1}}{sqrt{1}}=\frac{sqrt{1}}{sqrt{-1}}$ $sqrt{-1}sqrt{-1}=sqrt{1}{sqrt{1}$ Hence $-1=1$

Fallacy 3: all angles are right angles

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

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