Write the mathematical formula for:

Square root of an odd number is odd (·, +, =, 1)

I am again having troubles with that type of task. I wrote thing like:enter image description here But again I don't really know if it is even close to the correct solution.

  • $\begingroup$ You have most of the right pieces, but your formula has an unquantified $z$, so it can't be right. It'd also be helpful for you to clarify your parentheses. $\endgroup$ – BallBoy Jan 25 '18 at 11:42
  • $\begingroup$ You mean that the $z$ on the right side is outside of a range of the $∃$ on the left side of implication? $\endgroup$ – James Smith Jan 25 '18 at 11:48

The syntax here is horrible! Double check your textbook for the rules for constructing well formed formulae. Also you have used "2" which isn't in your official vocabulary. But most basically, your quantifiers haven't got the right "scope", i.e. aren't governing what you need them to govern.

It greatly helps to take translations in stages, using "Loglish" as a half way house as you go along. So you want to render

The square root of an odd number is odd

And you understand that as a universal generalization, so

(For any number $x$)(if $x$ is the square root of an odd number, then x is odd)

Which unpacks as

(For any number $x$)(if there is an odd number $y$ where $x \cdot x = y$, then $x$ is odd)

Let's now make things simpler by explicitly assuming that the domain is (natural?) numbers, so we needn't explicitly type the variables. That seems to be assumed in the question, as you aren't given a symbol for "is a number". Then we have, more simply

(For any $x$)(if (there exists $y$)($y$ is odd and $x \cdot x = y$), then $x$ is odd).

Now putting in the quantifier, conjunction and conditional symbols we get

($\forall x$)(($\exists y$)($y$ is odd $\land$ $x \cdot x = y$) $\to x$ is odd).

Note the bracketing. So now you just need to translate "$x$ is odd", "$y$ is odd" given your available apparatus.

"$w$ is odd" can be rendered something like this: $(\exists z)(w = (z + z) + 1)$

or equally

"$w$ is odd" can be rendered something like this: $(\exists z)(w = (1 + 1)\cdot z + 1)$

So now plug in this sort of thing, and you are done! Though your official syntax may not need brackets round quantifiers, and I've been a bit sloppy with brackets in the last two formulae (where?).

Now, I'm not suggesting that you write down all the Loglish mash-ups -- but it is worth writing down some when faced with this sort of translation excercse. Thinking in stages like this will keep you on track!

If your library has a copy of P-t-r Sm-th's Introduction to Formal Logic, check out the (short but quite good!) chapters 22- 24 for more along these lines.


Actually, the solution is quite simple.

First things first, we need to assume that $\sqrt{x}$ is an integer. Then, we can make the following inferences

  • An odd number times an odd number gives an odd number.

  • An even number times an odd number gives an even number.

  • An even number times an even number gives an even number.

These then means that for a square number $x$ to be odd, it's square roots must be odd.

  • $\begingroup$ The question can be restated as: "Using the formulas of predicate logic, write a mathematical formula for the statement: The square root of an odd number is odd." $\endgroup$ – Rohan Jan 25 '18 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.