Question

Consider the following argument. If there are as many rational numbers as there are irrational numbers, then the set of all irrational numbers is infinite.

The set of all irrational numbers is infinite. ∴
There are as many rational numbers as there are irrational numbers. Let p = "there are as many rational numbers as there are irrational numbers," and let q = "the set of all irrational numbers is infinite."
Is the argument valid or invalid?
Choose the answer that shows the symbolic form of the argument and justifies your conclusion.
a. form: q ∧ p valid, specialization p ∴ q
b. form: p → q invalid, converse error q ∴ p
c. form: q ∧ p invalid, converse error p ∴ q
d. form: p ∨ q invalid, inverse error q ∴ p
e. form: q → p valid, generalization p ∴ q

Answer 1

The argument commits the converse error, mistaking the statement 'if p then q' to also mean 'if q then p', which is logically invalid.

The correct symbolic form of the argument and justification is b: form: p → q invalid, converse error q ∴ p.

Step-by-step explanation:

The argument presented is a logical form which asserts that if p implies q, and q is known to be true, then p must be true. However, this is an instance of the converse error or affirming the consequent, which is a common logical fallacy. The argument's form does not ensure that the conclusion follows necessarily from the premises, thus the argument is invalid.

Looking at the options provided, the correct symbolic form of the argument and justification is b: form: p → q invalid, converse error q ∴ p. This means that even though the set of irrational numbers is infinite, it does not logically follow that there are as many rational numbers as there are irrational numbers solely based on the infinity of the latter.

Answer 2

b. form: p → q invalid, converse error q ∴ p

Step-by-step explanation:

Let

p = "there are as many rational numbers as there are irrational numbers,"

let

q = "the set of all irrational numbers is infinite."

We can write the above argument as:

If there are as many rational numbers as there are irrational numbers, then the set of all irrational numbers is infinite

p → q

The set of all irrational numbers is infinite

q

There are as many rational numbers as there are irrational numbers

p

So collectively it can be written as:

p → q

q

∴p

So it looks like

if p then q

q

therefore p

This is a fallacious argument. This means the conclusion can be false even when the premises are true. This type of argument is invalid.

Suppose we know that “If there are as many rational numbers as there are irrational numbers, then the set of all irrational numbers is infinite” is a true conditional statement. We also know that "the set of all irrational numbers is infinite" is true. This is not enough to say that "there are as many rational numbers as there are irrational numbers," is true. The reason for this is that there is nothing logically about “If p then q” and “q” that means p must follow. So this is a converse error and since converse error is an invalid method of inference rule so this argument is invalid.

Let us prove this argument is invalid with a truth table:

p q p → q q p

T T T T T

T F F F T

F T T T F

F F T F F

Since we know that the premises are:

p → q and q

and the conclusion is p

and an argument is valid if and only if all of its premises are true, then the conclusion is true. We should check that whenever both p → q and q are true then p is true but the third row fails. Thus this is an invalid argument.