• Ulrik
    21
    Dear philosophers,

    I am thankful for your replies on my other syllogism thread. Here's another reasoning that I fail to see. This time it veers off into a more mathematical direction. Here goes:

    If you square an odd number, then the number will be odd
    If you square an even number, then the number will be even

    Therefore

    If the square of a number is even, the number must be even

    I thought this reasoning was invalid, but it is valid. I thought it was invalid because it seemed a lot like affirming the consequence to me. If you focus on the 2nd premise, you could translate it to:

    If p, then q

    The conclusion says:

    If q, then p

    Therefore the conclusion is invalid.

    I drew the following diagram to attack the problem:

    The little 2 means squared.

    Engish-version-number-deduction.png

    So what we see here is that if you start with a squared number as in the conclusion, you cannot be certain that the squared number is a part of the even domain, because there might be numbers that are not even or uneven. Nowhere in the premises it says that there are only even or uneven numbers. I have bracketed all my knowledge of mathematics and focused only on the premises. Or so I thought. Gosh I'm confused. I need a break.
  • Pair o'Ducks
    6
    I agree that the conclusion is invalid on the basis of these two premises alone. For it to be valid, you would need to add the premise that all numbers are either even or odd (which, of course, is not the case). Or you could adjust the conclusion to fit the demarcation that the premises imply: If the square of an integer is even, that integer must be even.
  • fdrake
    2.7k
    'If the square of a number is even, then that number must be even' - this is true. But does it follow from the premises alone that:

    (1) If a number is even, then its square is even.
    (2) If a number is odd, then its square is odd.

    ?

    Well, in this set up, we don't know anything about the relationship of odd and even, and we don't know anything about prime factorisations or that even means 'is divisible by 2 with no remainder'... The only premise here which is even related to even numbers and squares of even numbers is (1).

    So the question becomes, can we conclude the statement: 'If the square of a number is even, then that number must be even' from the statement 'if a number is even, then its square must be even'? No. You're absolutely right to say that this is affirming the consequent. Affirming the consequent is an invalid argument of the form (P implies Q, assume Q, therefore P).

    Setting it up like it appears in the Wikipedia article - our 'P' is 'a specific number is even', our 'Q' is 'the square of the number in P is even'. So (1) translates to 'P implies Q' or equivalently 'P=>Q'. Now we're tasked with arriving at the conclusion P using only the assumption Q... And we can't.

    What this example highlights is the difference between an argument failing to establish a conclusion since it is invalid, and that conclusion being true or false. The take home message of this is that arguments are attempts to link premises to conclusions, a valid argument is a link that transmits the truth of premises to the truth of the conclusions, but it can still be the case that an argument makes a true conclusion by means of invalid reasoning. We can also arrive at a false conclusion by means of valid reasoning - just when our premises are false.

    Validity has the technical meaning of 'an argument is valid if the truth of its premises ensures the truth of its conclusions', which does not guarantee that the conclusions of a valid argument are always true, nor that we cannot arrive at false conclusions through a valid argument. All validity requires is that if the premises are true then the conclusions are true.
  • fdrake
    2.7k
    Just for reassurance.

    Nevertheless, it is true that 'if the square of a number is even, then that number is even', and here is a different valid argument (though not presented in the strict language of a formal reasoning system) to show it:

    (A) Every number has a prime factorisation. (assumption, true)
    (B) If a number, x, is equal to another number, y, squared; then the prime factorisation of x must consist solely of all the factors of y squared. (assumption, true)

    (so if we have x = 144 = 12^2, then y = 12, the prime factorisation of x=144 is 12*12=3*4 * 3*4 = 3*3*2*2*2*2=3^2 * 2^4, then the prime factorisation of y must be (3*2)^2)

    (C) If a number contains 2 as a prime factor, then that number is even and vice versa (assumption, true)
    (D) If x is the square of a number y and is even, then y must also be even (conclusion, true, from A,B,C and since the argument is valid)

    For interest, the difference between something being true (or false) and there being a valid argument which demonstrates that truth (or falsity) has really deep consequences in logic and mathematics.
  • SophistiCat
    838
    Well, in this set up, we don't know anything about the relationship of odd and even, and we don't know anything about prime factorisations or that even means 'is divisible by 2 with no remainder'... The only premise here which is even related to even numbers and squares of even numbers is (1).fdrake

    Either you understand what is written, or you don't. If you don't, then nothing more can be said. If you do, then you know what "odd" and "even" mean. (But then if you do, you don't really need to go through this logical exercise in order to prove the conclusion - you could prove it by other means.)

    I thought it was invalid because it seemed a lot like affirming the consequence to me. If you focus on the 2nd premise,Ulrik

    So the question becomes, can we conclude the statement: 'If the square of a number is even, then that number must be even' from the statement 'if a number is even, then its square must be even'?fdrake

    Why are you focusing on the second premise alone when two premises are given?
  • TheMadFool
    4.1k
    I thought this reasoning was invalid, but it is valid.Ulrik

    It is invalid.

    However, if one were to use math then we have
    x=even number = 2n

    The square of every even number would be even of the form 4n*n
    So imagine any even square number and take its square root squareroot of 4n*n = 2*n which is even.

    So one of the statements should read "The square of a number is even IF AND ONLY IF the number is even"
  • fdrake
    2.7k
    Either you understand what is written, or you don't. If you don't, then nothing more can be said. If you do, then you know what "odd" and "even" mean. (But then if you do, you don't really need to go through this logical exercise in order to prove the conclusion - you could prove it by other means.)SophistiCat

    For the purposes of a logic exercise all knowledge that you bring to an argument you're analysing has to take the form of stated propositions. You're usually given the stated propositions, and no more. Reasoning 'out in the wild' does not have this restriction, it's more free form and you can bring whatever you like to bear on the problem you're tackling. Whereas all the propositions and the rules you can use to link them (like the rules in the sequent calculus or propositional logic) are stated up front in the exercise.

    Edit: in case this isn't clear, a good example here is it's very often required in logic exercises that you can't use 'theorem introduction' rules, so if something like de Morgan's laws or the equivalence between disjunctive syllogism and material implication are not axiomatic inference rules of the system, you can't use them.
  • Ulrik
    21
    Thank you Pair o'Ducks, fdrake and Sophisticat!

    I had other duties the last few days hence my late reply. It sure is reassuring that the reasoning is actually invalid, though the book says it is valid. I found it to be invalid precisely because of reasons mentioned in this thread. The book probably did mean integers, if integer implies : 'there are no other numbers other than even or odd'. In that case, if I'm not mistaken, the total number of possible even numbers is equal to the possible number of even numbers squared, hence it is not possible to not find a number that is even that does not have a square that is also even, which is a long way of saying:

    If a number is even, if you square it, it will be even
    and
    If you have an even squared number, if you take the root of it, it will be even

    Regarding SophistiCat's concern of focusing only on the 2nd premise: I did use the 1st premise in my notes, which I didn't write out all here.

    In any case thanks again and I am now ready to move on to the next exercise!
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