Common mistakes

As humans, we all make mistakes. To avoid typographic errors, we must proofread the script we write. We must carefully verify each algebraic manipulation in proof to ensure correctness. Overall, the proof must be a valid argument type, and every deduction must be according to inference rules. Sometimes, errors in algebraic manipulation are straightforward to catch, but sometimes they are very subtle. Let’s look at the following example of a wrong derivation with a subtle mistake:

1.x=y2.x2=xyMultiplying both sides with x.3.x2y2=xyy2Subtracting y2 from both sides.4.(x+y)(xy)=y(xy)Factorizing.5.x+y=yCancelling (xy) from both sides.6.2y=ySubstituting from line 1.7.2=1Cancelling y from both sides.\begin{aligned}{} &1.& x &= y\\ &2.& x^2 &= xy &\text{Multiplying both sides with }x\text{.}\\ &3.& x^2-y^2 &= xy -y^2 &\text{Subtracting }y^2\text{ from both sides.}\\ &4.& (x+y)(x-y) &= y(x-y) &\text{Factorizing.}\\ &5.& x+y &= y &\text{Cancelling }(x-y)\text{ from both sides.}\\ &6.& 2y &= y &\text{Substituting from line 1.}\\ &7.& 2 &= 1 &\text{Cancelling }y\text{ from both sides.} \end{aligned}

We notice that every step is fine except the fifth step if we look closely. As x=yx=y, we can not cancel out (xy)(x-y) from both sides because it equals zero. Such mistakes can lead to inconsistent results, as in line number seven.

Now let’s look at some common mistakes of argument.

The fallacy of affirming the conclusion

Let’s assume we have x(P(x)Q(x))\forall x (P(x)\Rightarrow Q(x)) is true for some propositions PP and QQ for a given domain. Instead of assuming P(a)P(a) true, we assume that Q(a)Q(a) is true for some element aa of the domain. The fallacy of affirming the conclusion is a wrong argument type as follows:

x(P(x)Q(x))Q(a)P(a)\begin{aligned}{} &\forall x (P(x) \Rightarrow Q(x))\\ &Q(a)\\ &\rule[0 pt]{80 pt}{0 pt}\\ &\therefore P(a) \end{aligned}

Let’s look at a few examples to clarify this error.

Example

Assume the domain set contains all the mammals.


Consider the following propositions:

  • C(x)C(x): xx is a cow.
  • M(y)M(y): yy is a mammal.

Assume the following statement as a fact:

xC(x)M(x).\forall x C(x) \Rightarrow M(x).

The above statement is saying that every cow is a mammal. Now, let’s look at the wrong argument.

As Frankie is a mammal, therefore, Frankie is a cow.

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