We have discussed various argument schemes. Since the argument is an ordered list of statements, each of which is either one of the premises or "derivable" from the combination of some subset of the preceding statements and one or more axiom, errors are possible in this organization.
Here we consider invalid arguments arising from errors made in organizing p Þ q.
1. Petitio Principii or Circular Reasoning
Suppose that the conclusion is assumed in the premise. Then p Þ q is true and the proof is valid, but it has not established the truth of q because it was already assumed as true.
This error is not rare with newspaper columnists and politicians. Here is one: "We need public transportation because everyone needs it."
(Note that since they have to make a lot of statments, this type of error will occur several times in the career of a politician or columnist - even if they do not make the mistake with any greater frequency than another person.)
Assuming the conclusion as part of the premise is easy to detect in the reading and so it seldom occurs in mathematics proofs.
However, it is not so easy to detect if the conclusion steals in as an assumption later on in the proof. This is how it happens.
Suppose, in showing that p Þ q is true,
we use a theorem or statement, say r, that itself uses q as its premise.
The diagram then would represent p Þ q. The red portion of the diagram is circular reasoning.
Thus p Þ q itself becomes circular reasoning because a part of it is circular. An example of this is the proof found in calculus books of the identity
| The circular part: r Þ q and q Þ r |
The subtle error  is that every proof makes use of an arc length of a circle as either the supremum or the infemum of the lengths of polygonal approximations, depending on whether an inscribed or escribed polygon is used; but this assumption amounts to assuming the identity we are trying to prove, for the identity implies the limit of the approximation.
2. Irreversible Argument
The error, which I have named Irreversible Argument, consists in concluding that p is true when p Þ q is true and q is true. Note that according to Modus Ponens, if q is true and q Þ p is true, then p is true. So if the argument p Þ q can not be reversed, we can not conclude that p is true.
Here is an example.
Prove that the cube root of 3 is greater than the square root of 2.
31/3 > 21/2
If 31/3 > 21/2 is true, then
(31/3)6 > (21/2)6.
Using the rule of exponents that (ab)c = a(bc), we get 32 > 23. That is, 9 > 8, which is clearly true.
This is not a proof that 31/3 > 21/2. However, since all its steps are reversible, let us place the steps in reverse order - that is work backward. Aha! There is an idea. In many problems to prove identities and inequalities, and even in other problems, an attempted proof might work backward to give the correct proof. Here then is our correct proof.
9 > 8 is true.
So 32 > 23.
So (31/3)6 > (21/2)6, by rule of exponents that (ab)c = a(bc).
Then ((31/3)6)1/6 > ((21/2)6)1/6, since a > b implies ac > bc for positive a, b, and c.
So 31/3 > 21/2. QED
The Arithmetic Mean - Geometric Mean Inequality is this: (a+b)/2 ≥ √(ab), for non-negative numbers a and b.
Consider this attempted proof:
All the steps are reversible, so reverse them and we get a correct proof.
The above examples show that working backward is a method of discovering proofs, but one must use it with caution, namely that the steps in the forward process must be reversible.
The e-d proofs in highschool and freshman university are also a hiding place of the error of Irreversible Argument.
We have seen that the error of Irreversible Argument is avoided by writing the steps in reverse order if every step is reversible. So we ask: what type of steps are irreversible?
A general rule is that the following operations might give an irreversible step:
(b) Multiplying by a term containing variables
The solution obtained by squaring an equation may not equal the solution of the original equation.
For, a = b implies a2 = b2, but a2 = b2 only implies that either a = b or a = -b.
Solve (√(x+3)) - 1 = x.
√(x+3) = x + 1
Squaring, we get: (x+3) = (x+1)2, which gives x2 + x - 2 = 0, or (x-1)(x+2) = 0.
The solution is x = 1, -2.
But is it?
Only x = 1 satisfies (√(x+3)) - 1 = x.
The other value of x stole in when we executed the irreversible step of squaring. x = -2, in this solution, is called an extraneous root or extraneous solution which is discarded after it is found that it does not satisfy the original equation. Notice that -2 satisfies the equation √(x+3) = -(x + 1); that is, it is the other - "extraneous" - solution of (x+3) = (x+1)2.
The moral is: Squaring must be used with caution in proofs.
Multiplying by a term containing variables:
If a = b, then ac = bc. However, ac = bc does not necessarily imply a = b. This problem occurs when the value of c can take on the value zero; for, a(0) = b(0), for any a and b.
Now, the same problem occurs if instead of multiplying by zero, we multiply with an expression containing variables that can take values that would make the expression evaluate to zero.
In fact, care should be taken in dividing with such expressions as well.
We have considered circular reasoning and the irreversible argument which are the main errors. We shall now look at other aspects that can make a proof invalid.
 F. Richman, A Circular Argument, The College Mathematics Journal (Mathematical Association of America), Vol 24, No.2, March 1993
G. Rubin Thomas
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