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My professor asked us to assess the validity of the following argument:

Some rational numbers are powers of 5. All integers are rational. Therefore, some integers are powers of 5.

My professor went back and forth on the validity of the argument after I questioned his logic. Finally, he asserted that the argument is invalid and he gave this argument:

Now let us abstract this argument, letting R represent the set of rational numbers, P the set of powers of five, and G the set of integers. Then the premises and conclusion become: 1) Some element of R belongs to P. 2) All elements of G belong to R. Conclusion: Some element of G belongs to P.

As a model of this argument, let R = { a, b, c }; P = { c }; and G = { a, b }. Are the premises true? 1) Some element of R belongs to P. 2) All elements of G belong to R. Does the conclusion now follow: Some element of G belongs to P?

I've spoken to another professor who says the argument is valid and I've seen different answers to this problem including this one:

from some textbook - name unknown

Link here: see exercise 6

So I ask, what is the truth?

EDIT:

In addition, for any open sentence P(x), is $$\exists x \in \mathbb{Q} P(x)$$ not equivalent to, $$\exists x (x \in \mathbb{Q} \Rightarrow P(x))$$ and if these are not equivalent, then why does the author, in the example below, rephrase the following quantified statement as an implication? enter image description here

EDIT 2: This post here is similar to my last question.

thank you.

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  • $\begingroup$ They are arguing that the conclusion follows from the premises? But it clearly does not. Consider this argument.... $\#1:$ Some rationals are between $\frac 12$ and $\frac 34$. $\#2:$ All integers are rationals. Therefore, some integers are between $\frac 12$ and $\frac 34$, How does this differ logically from what you were shown? $\endgroup$
    – lulu
    Commented Feb 13, 2018 at 0:13
  • $\begingroup$ It's confusing because of course we know that some integers, such as $5$, are powers of $5$. So the conclusion is true in the abstract. But as a matter of pure logic, I don't understand the reasoning shown here. $\endgroup$
    – lulu
    Commented Feb 13, 2018 at 0:14
  • $\begingroup$ Some are arguing that the argument is invalid others argue that it is valid. I am confused as well. $\endgroup$ Commented Feb 13, 2018 at 0:17
  • $\begingroup$ Well, unless I am missing something the argument seems entirely invalid. It's like saying "some good meals are vegetarian. Every steak dinner is a good meal. Therefore some steak dinners are vegetarian." $\endgroup$
    – lulu
    Commented Feb 13, 2018 at 0:20
  • $\begingroup$ Regarding my very last question. My professor asserts that they are not equivalent but then I have this other textbook saying that they are equivalent hmmm so that adds to my confusion $\endgroup$ Commented Feb 13, 2018 at 0:20

2 Answers 2

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The argument is invalid. Here is a refutation by logical analogy:

Some coins are dimes

All nickels are coins

Therefore, some nickels are dimes

The argument based on formal logic notation fails, since it uses the wrong symbolizations. For example, some rational numbers are powers of five needs to be symbolized as:

$$\exists x (Q(x) \land R(x))$$

and not as:

$$\exists x (Q(x) \rightarrow R(x))$$

So ... either the text was asking you to find the error in the 'Solution' ... or the text provided a horribly mistaken Solution! Given how everything else labeled 'Solution' seems to be treated as the actual answerk to the exercises, I fear it's the latter .. what text is this?!

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  • $\begingroup$ I fear the same too. I made an edit to my post, hopefully, you can clarify it. Thank you. $\endgroup$ Commented Feb 13, 2018 at 2:55
  • $\begingroup$ @rhoward The universally quantified statement can indeed be written as a conditional, but not the existential. That is: $\forall x \in S: P(x)$ can indeed be written as $\forall x (x \in S \rightarrow P(x))$, but $\exists x \in S : P(x)$ rewrites as $\exists x (x \in S \land P(x))$ $\endgroup$
    – Bram28
    Commented Feb 13, 2018 at 2:57
  • $\begingroup$ I've been trying to find the name of text too. The link is all I have. $\endgroup$ Commented Feb 13, 2018 at 3:05
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The argument:

"Some rational numbers are powers of 5. All integers are rational. Therefore, some integers are powers of 5."

is not valid

The rationals which are powers of 5 are not necessarily integers. For example $\frac {1}{32}$ is a power of 5 but it is not an integer. All integers are rational does not mean all rationals are integers.

If all rationals were powers of 5 then we could argue that all integers are powers of 5, but from some rationals are powers of 5 we can not argue that some integers are powers of 5.

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