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I'm having some issues understanding predicate logic. For example, given the proposition:

These propositions are based on a set of numbers.

$$\forall x \exists y (x \cdot y = x)$$

This means for every $x$ there exists a $y$ that when you multiply you get the same $x$ as an answer. This is true.

However the next statement is

$$\exists y \forall x (x \cdot y = x)$$

This means there exists a $y$ for all $x$'s where when you multiply that $x$ by $y$ you get $x$. I think this is also true (the number 1)

However, I am under the impression that this is wrong and I'm missing something. Could anyone help me understand better?

Rodrigo de Azevedo
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denn1s
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  • I really can't say more rather than, as real number form commutative group, $y$ is unique for all besides $x=0$. (for first statement if $x=0$, there are many $y$-s, for example. For others the $y$ is indifferent of what you $x$ is and so is the $y$, one in the second statement, which takes the same $y$ for all $x$-s). – kolobokish Nov 16 '16 at 17:37
  • Your interpretation is right, and both statements are true. However, the big difference is that the second statement implies that there is a single $y$ that works for all $x$ simultaneously, while the first statement does not. – Arthur Nov 16 '16 at 17:37

1 Answers1

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The first statement says that, for every number $x$, you can find a $y$ that will act as multiplicative identity for it. It could be a different identity for every $x$.

The second statement says that there is a number that acts as multiplicative identity for every other real number. There is indeed such a number, and we call it $1$.

Does that help?

G Tony Jacobs
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