by

**Sampad Sengupta**A fallacy is a mistaken belief, one based on unsound argument. It may be an idea that a lot of people think is true but is in fact false. Now, in mathematics, a proof is an argument, a deductive argument for a mathematical statement. It may use statements which are already established, i.e. theorems, and logic; a proof can be made by mathematical induction, contradiction, exhaustion, and many other ways. However, even some arguments like these may turn out to be fallacies. There are certain so-called mathematical proofs which may apparently show some unexpected statements to be true by cleverly concealing areas in the proof, but these statements are indeed false.

Here are a
few examples of such fallacies and the truth behind each one of them:

**1)**

**Proof of**

*2=1*:**Let**

*a*and*b*be equal non-zero quantities,*a = b*

**Multiplying through by**

*a**a*

^{2}= ab**Subtracting**

*b*from both sides^{2}*a*

^{2}- b^{2}= ab-b^{2}**Factorising the equation**

*(a-b)(a+b) = b(a-b)*

**Dividing out**

*(a – b)*
a

*+b = b***Since**

*a=b**b+b = b*

*2b = b*

**Dividing both sides by**

*b**2 = 1*

At first glance this looks quite remarkable, but if
you look closely, you can spot the error. The proof starts off soundly, but
then the mistake creeps in when we divide out by

*(a-b)*. Since*a*and*b*are both equal,*(a-b)*is zero. Since division by zero is undefined, the argument is invalid. If it were valid, then it would be possible to prove any number to be equal to any other number, but it isn’t.**2)**

**Proof of**

*0=2*:*0 = 1-1*

**Since square of 1 = 1**

*0 = 1-1*

^{2}*0 = 1 - √1*

^{2}**Adding and subtracting 4 under square root as they will cancel each other out**

*0 = 1 - √1*

^{2}– 4 + 4**Since**

*(1*^{2}– 4 + 4) = (1 – 2)^{2}*0 = 1 - √(1-2)*

^{2}**We then remove the square root and multiply out the brackets**

*0 = 1 - (1 – 2)*

*0 = 1 – 1 + 2*

**The two**

*1*s cancel out leaving*0 = 2*

Once again, an equation like this cannot be true.
The problem lies when we are taking the square root of 1

^{2}. Taking square roots requires the use of the double plus-or-minus sign__+__(or absolute values). In this case, when we take the square root of 1^{2}, the results should be both +1 and –1. Thus, this proof is also invalid.
3)

**Proof of**(this time by a different method)*0=2*:**Let us take an equation which we know to be true,**

*cos*

^{2 }*x = 1 – sin*

^{2 }*x*

**Taking the square root of both sides**

*cos*

*x = (1 – sin*

^{2 }*x)*

^{1/2}**Adding**

*1*to both sides*1 + cos*

*x = 1 + (1 – sin*

^{2 }*x)*

^{1/2}**By evaluating this when**

*x*= 180^{0}*1 – 1 = 1 + (1 – 0)*

^{1/2}**Thus showing**

*0 = 2*

In equations like these, the fallacy may be
concealed effectively in notation.
Similar to the previous one, the error in each of these examples
fundamentally lies in the fact that any equation of the form

*x*^{ }^{2}*= a*has two solutions:^{2}*, provided*__+__a*a*is not equal to 0.
In this example, only when the square root of

*cos**x*is positive is the equation valid, but when*x*is set to 180^{0}, the proof is invalid.
4)

**Proof of***1=2*using complex numbers and the imaginary unit (*i =*√-1 ):*- (1/1) = - (1/1)*

*- 1/1 = 1/-1*

**Square root both sides**

*√-1/1 = √1/-1*

**Simplifying**

*√-1 / √1 = √1 / √-1*

**Converting √-**

*1*to imaginary number

*i**i / 1 = 1 /*

*i*

**Multiplying both sides by ½**

*i / 2 = 1 / (2*

*i)*

**Adding**

*3/(2*

*i)*to both sides*i/2 + 3/(2*

*i) = 1/(2*

*i) + 3/(2*

*i)*

**Multiplying both sides by**

*i**i(*

*i/2 + 3/(2*

*i)) =*

*i(1/(2*

*i) + 3/(2*

*i))*

**Simplifying**

*(*

*i*

^{2})/2 + (3*i)/2*

*i =*

*i/(2*

*i) + (3*

*i)/2*

*i*

**Converting**

*i*term to^{2}*-1*and adding fractions*(-1)/2 + 3/2 = 1/2 + 3/2*

*1 = 2*

This so-called proof makes use of complex numbers,
thus making it hard to find the error. The
fallacy here lies in the third step. In
simplifying from step two to three, we try to make two things equal that are
not. We have √(-1/1) = √(1/-1) in step
two. The left-hand side (LHS) does
simplify to √(-1)/√(1), which is the LHS in step three. The right-hand side (RHS) in step two √(1/-1)
does not simplify to √(1)/√(-1), which is what we see as the RHS in step
three. In fact, it is -√(1)/√(-1). A more careful examination of negative
numbers will help explain the fallacy clearly. The product of (-1)(-1) is one. So √(-1*-1) =
√(1) = 1. Yet √(-1)√(-1) =

*i***i*=*i*^{2}= -1. So these two products are not the same. Hence the rule of √(ab) = √(a)√(b) holds true as a rule when a and b are both positive numbers. For negative and complex numbers, however, this rule fails to hold true.
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