I faced a question in the book "Discrete Mathematics" by Rosen. The question is this:

Determine whether the argument is valid:

If $n$ is a real number with $n > 3$, then $n^2 > 9$ : Suppose that $n^2 \le 9$. Then $n \le 3$.

I really think that this argument is valid by using modus tollens, but the answer mentioned at the end of the book is this: "Fallacy of begging the question".

Can someone explain this for me? Thanks.

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    $\begingroup$ It seems valid to me. $\endgroup$ – Shaun Mar 25 '17 at 9:10
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    $\begingroup$ @shawn The conclusion is true but the argument is invalid. $\endgroup$ – MJD Mar 25 '17 at 9:17
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    $\begingroup$ It's not so much that the argument is invalid, it's just that it's not an argument. $\endgroup$ – 5xum Mar 25 '17 at 9:26
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    $\begingroup$ In the quest to clarify the other answers without creating yet another near-duplicate, here's what I consider the clearest statement of the issue: The sentence with the argument is poorly worded. It reads as though it were saying, "If P then Q. Not Q. Therefore not P." That would obviously be valid. However, the colon indicates that the argument is actually found in the second half of the sentence and runs: "P (n^2 <= 9). Therefore Q (n<= 3)." It is invalid because we are missing the premise "if P then Q." (The first half of the sentence is just an equivalent restatement of this argument.) $\endgroup$ – Luke Sawczak Mar 25 '17 at 23:11
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    $\begingroup$ From the wording, it's not clear what the hypotheses or the conclusions of this supposed argument are. So it's "not even wrong." $\endgroup$ – user360874 Mar 26 '17 at 1:41

10 Answers 10


It is not clear from the phrasing in your quote whether the statement "If $n$ is a real number with $n > 3$, then $n^2 > 9$" is meant to be a given premise or the statement that we're trying to prove.

If it's a premise, and we're trying to use it to prove that $n \le 3$, given the additional assumption that $n^2 \le 9$, then the argument is indeed correct: given $A \implies B$, we can deduce $\lnot A$ from $\lnot B$.

If, instead, the first statement in the quoted argument is supposed to be the theorem that we're trying to prove, then the proof is invalid. While we could indeed deduce $n > 3 \implies n^2 > 9$ from $n^2 \le 9 \implies n \le 3$ (those two statements being logically equivalent), the quoted argument contains nothing that could be used to prove $n^2 \le 9 \implies n \le 3$ in the first place, unless we incorrectly assume the statement that we're trying to prove (or something else that has not been explicitly stated).

Presumably, the book you're reading explains how such arguments given in it are to be parsed. In particular, if the arguments in the book are written in the format "Theorem to be proved: Proof of theorem," then this particular argument is invalid; if they're instead meant to be parsed as "Established premises: Assumptions. Result," then it's valid.

Unfortunately, while there exist several established conventions like this, none are quite universal enough to be safely assumed without knowing which convention the author follows. When writing proofs yourself, to avoid such confusion it's usually a good idea to clearly and explicitly indicate which statements are axioms or prior theorems, which are temporary assumptions, and which are the new theorems that you're trying to prove.

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    $\begingroup$ We can infer from the book's answer that the second interpretation is intended. $\endgroup$ – Daniel R. Collins Mar 26 '17 at 13:25
  • $\begingroup$ We can infer from the book's misprint that the first interpretation is the intended one. The Ex is in Ch.1.6: Rules of Inference, sub-chapters: Valid arguments in Propositional Logic, Rules for PL (e.g. Modus Tollens), Using Rules to Build Arguments. $\endgroup$ – Mauro ALLEGRANZA Mar 28 '17 at 8:04

It's not really a proof, that's the problem. It's just rephrasing the statement and then claiming that it's obvious. If the arguer actually showed that if $n^2 \leq 9$ then $n \leq 3$, it would be a valid proof by contraposition.

It's like saying: "We prove Fermat's Last Theorem. Indeed, let $n \geq 3$. Then $x^n + y^n = z^n$ is equivalent to $y^n + x^n = z^n$, which never happens, so we're done.".

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    $\begingroup$ I agree, all is revolving around two things : applying the contraposition method is a thing, but proving the starting point is the other. Both have to be statisfied for a valid proof. $\endgroup$ – zwim Mar 25 '17 at 9:04
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    $\begingroup$ @Patrick Stevens actually, we are not looking for a proof, we want to know whether this argument is valid or not. and i think these two are different. $\endgroup$ – Fatemeh Karimi Mar 25 '17 at 10:54
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    $\begingroup$ The argument is valid, but it doesn't argue anything. $\endgroup$ – Patrick Stevens Mar 25 '17 at 12:38
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    $\begingroup$ @FatemehKarimi: Please see my answer for more precisely why the stated argument should be considered invalid under any reasonable interpretation, especially in mathematics, which you have expressly referred to. Furthermore, your book is an introductory book and so one expects all arguments to be properly justified. Once you can see that there is no essential difference between your quoted argument and the similar-looking argument I gave in my answer, you would see what's wrong. $\endgroup$ – user21820 Mar 25 '17 at 15:31

I think this may just be a (mutual) misunderstanding between you and the textbook of the symbol ":" in the actual textbook question.

The textbook meaning of ":" seems to be "Because".

Taking that as the ":" meaning, the argument is:

If n is a real number with n>3, then n^2>9 (because,) if n^2≤9, then n≤3.

So, if the second part (if n^2≤9, then n≤3) is supposed to be an argument in the favor of the implication n>3==>n^2>9, it actually is not, as it is just re-stating (begging) the question.

While you (and, mind you, I as well) would read ":" as "therefore", making it a legitimate modus tollens:

If n is a real number with n>3, then n^2>9 (therefore,) if n^2≤9, then n≤3.

Edit: My apologies for not noticing - this is actually what @Ilmari Karonen wrote in his earlier answer.


It is only a misprint.

See page S-9 of 7th edition (2013): there are answers for questions 19 a), b), c) and d), where Exercise 19 (Ch.1.6, page 79) has only three questions.

Your doubt is relative to question:

19 b) If $n$ is a real number with $n > 3$, then $n^2 > 9$. Suppose that $n^2 ≤ 9$. Then $n ≤ 3$.

The answer says:

19 c) Valid argument using modus tollens.

In 5th ed (2002), we have (Ex.13, page 75) four cases, where the 19 b) of 6th and 7th editions is numbered c); in this edition we have an additional b) (deleted from successive editions):

The number $\log_2 3$ is irrational if it is not the ratio of two integers. Therefore, since $\log_2 3$ cannot be written in the form $a/b$...

that is clearly an example of circular reasoning (aka begging the question).

The argument:

$$ \cfrac{p \to q \ \ \ \lnot q}{\therefore \lnot p} $$

is clearly valid.


The claim is of the form:

If A then B.

The argument is of the form:

If not B, then not A. [Therefore if A then B.]

The argument is invalid because not every step is justified. Worse still, the argument is not even of a sound form because it relied on nothing about A and B, and hence can be used to prove:

If $n \le 1$ then $n^2 \le 1$.

You would object to this, saying, "Why does assuming $n \le 1$ allow you to deduce $n^2 \le 1$?" And I would object exactly the same way to the invalid argument:

Why does assuming $n^2 \le 9$ allow you to deduce $n \le 3$?

That is ultimately why the textbook classifies this as (essentially):

Begging the question.

Since it's much harder to justify the unjustified deduction than the rule that from "not B implies not A" one can deduce "A implies B". Often in mathematics one does not even bother to justify purely logical inference rules such as these.

  • $\begingroup$ I hope you do realize that "If $n \le 1$ then $n^2 \le 1$." is an invalid statement, where $n$ is given to be a real number, just as in your quote. $\endgroup$ – user21820 Mar 25 '17 at 15:34

There are actually two important details which make this argument invalid. One of them has been pointed out in the previous responses, but the other has not. Namely, we need to read the word "with" as a logical "and" operator. In symbolic logic we now have the following:

(A ^ B) -> C : Suppose ~C. Then ~B

What has been pointed out is that "Suppose ~C" isn't derived from the initial premise "(A ^ B) -> C". This means that "~B" isn't a conclusion which follows from it's premises. In fact, the second premise here is "Suppose ~C. Then ~B." This is why the argument is a fallacy of the form "begging the question."

What has not been pointed out is that even as a premise, "Suppose ~C. Then ~B." is falsely derived. This is easily shown by applying DeMorgan's Law in the contrapositive:

~((A ^ B) -> C) <-> ~C -> (~A V ~B)

And given ~C, we would have to first establish A, "n is a real number" before we could conclude ~B.


Let's start with the statement: If n is a real number and $n > 3$, then $n^2 > 9$.

Seeing this statement, I can either say "Yes, of course, that's obvious". Or I can say "Well, it's not obvious, here is the proof".

Then in my proof I use the statement "Suppose that $n^2 ≤ 9$. Then $n ≤ 3$". With this statement, I can also either reasonabley claim it's obvious, or claim it's not. But it is exactly as obvious as the first statement, not one bit more or less obvious. So using the second statement in a proof of the first statement, without proving it, is pointless. Either the first statement was so trivial that it didn't need a proof. Or it wasn't trivial, then the second statement is also not trivial, so we don't have proof.

In addition, we can claim that both statements are the same. If you use a statement X in a proof for the statement X, that's called "begging the question", and nothing is proved. The two statements are not identical, but we can say that their meaning is identical, so this is "begging the question".


Modus Tollens is in the form:

$$P \rightarrow Q$$ $$\neg Q$$ $$\bullet P$$

But, you are only given:

$$P \rightarrow Q$$

You do not have the second premise of the modus tollens form (in this case, you do not know if $n^{2} \leq 9$ is true; you suppose it) to conclude.

  • $\begingroup$ Why the downvote? $\endgroup$ – Kyoma Mar 26 '17 at 0:53

Logically, it's valid. The other things aren't mathematics and shouldn't be discussed in the mathematics context.

There are several problems if someone insists:

  • It has assumed it is obvious that "if $n^2 \le 9$, then $n \le 3$". But yes, it is obvious. In the proofs of most other statements, you wouldn't be bothered to explain why this is true. Unless you are just learning about how to prove inequalities like this. But if it is discussing modus tollens, I think it's natural to assume it is not the case.
  • By this argument, "if $n > 3$, then $n^2 > 9$ is obvious, too, which doesn't need such a proof. But nothing can stop someone adding this redundant step. Redundancy doesn't make it wrong. People don't usually write this kind of proofs because they don't tell any new information. But thinking in the mathematics way, they are not more wrong than "if $1 = 2$, then (whatever)".
  • The statement is too simple that could make people too shy to assume it obvious, and to write a blank proof. They may try to guess the asker's intention and write more details. But guessing it wrong doesn't make it "mathematically" incorrect, as the asker's intention is unlikely defined in mathematics in the context.
  • One cannot say it doesn't require any guesses when it is that obvious, and there isn't a context such as "theorems described in this book before this page". Should we use the theorem that multiplying a positive number on both side retains the inequality, like the people who still need to write such proofs, or use the fact $n^2$ is strictly increasing for positive $n$, to explain what is called "obvious" (but do we also prove it is strictly increasing?), or start with the basic axioms about real numbers, to make sure it is exactly correct?

Also, it's impossible to identify a "fallacy" mathematically in arguments like this. You can't distinguish most if not all named fallacies from the fallacy of "creating a theorem out of nothing, irrelevant to the reasoning". It could be answered using some common sense. But that's not mathematics.

Things like this might be good examples demonstrating what could go wrong if with enough explanations. But they are really bad exercises as the answers are not mathematics. One possible fix is to write down the exact theorems they have used in the questions, but that usually makes the answers too obvious (which also implies that, if someone supposedly cannot answer the question, the book is teaching people in the wrong way in the first place). Another way is just to make the conclusion wrong.

  • $\begingroup$ "If $n^2 \le 9$, then $n \le 3$" is only obvious if you're working with the real numbers, or a subset thereof. There are plenty of settings (such as modular arithmetic or the complex numbers) where that statement is either false or meaningless. $\endgroup$ – Ilmari Karonen Mar 26 '17 at 18:59

The problem is: What if $n \leq 3$ were true even when $n^2 > 9$?

You didn't prove that is impossible!

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    $\begingroup$ Why in the world is this downvoted..? $\endgroup$ – Mehrdad Mar 26 '17 at 9:20
  • $\begingroup$ I did not downvote, but this answer seems to be talking of the converse of the statement that needs to be proved. We need $n\gt3\implies n^2\gt 9$, not the other way around. $\endgroup$ – GoodDeeds Mar 26 '17 at 17:10

protected by Alex M. Apr 1 '17 at 16:20

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