Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just have a small question! Really basic I'm sure but something is bothering me. Take note of the following statement:

$\forall x \in I , \exists y \in I$ such that $xy \in I $

Does this statement say that for all $x$ in $I$, there exists a $y$ in $I$ or does it say that for every single $x$ in $I$, there exists a $y$ that makes the statement true.

So for example if we took $x1$ to be $\pi$ and $x2$ to be $\sqrt 2$, can we assign them a unique $y$ for each one or for both of them?

It is a very fine difference but in proving the statement, it is rather confusing. I understand that the statement is true however but would like to solidify my reasoning.

Thank you :)

Ah yes, I understand now! Thank you everyone :) Really appreciated!

share|cite|improve this question
The statement says "For all x in I there exists a y, that is in I, such that the multiple xy is also in I." It does mean ANY x, as long as x is in I. But remember, y also has to be contained in I for this statement to hold true. – Steven Walton Apr 11 '13 at 2:38
The $y$ depends of the value of $x$, that is, for $x_1$ you have a $y_1$ and for $x_2$ you have a $y_2$. – FASCH Apr 11 '13 at 2:40
up vote 4 down vote accepted

It means both: for all $x \in I$ there exists a $y \in I$ equivalent to "for every single x in I, there exists a $y$ that makes the statement true.

But what this means, with respect to your follow up questions, is that each $x$ can have some particular y (which may depend on that x) for which the statement is true. It does not mean that there must be one y that satisfies the statement for every x. So for any particular x, there can exist a particular y, such that the statement is true. As $x$ varies, so can the "some particular" y.

So for example,

$$\forall x \in \mathbb R,\;\exists y \in \mathbb R\;( x \lt y)$$ is true. For any $x$ there is some y that is greater than that $x$. If $x = 1/2$, then there exists a $y = 1$ suffices to make $x\lt y$ true. If $x = 4$, then there is some $y$, say $5$, for which $x\lt y$.

This is different that $$\exists y \in \mathbb R,\;\forall x\in \mathbb R \;(x < y)$$ which is false: there is no $y \in \mathbb R$ that is greater than every $x \in \mathbb R$.


You might appreciate this related post, which helps clarify the difference between the $\exists y \forall x$... and $\forall x \exists y$...You'll also find a list of posts linked to that post, which address the same question.

share|cite|improve this answer

In many contexts, $\forall x\in\mathcal{I},\,\exists y\in\mathcal{I},\,...$ means given an $x$ there is a $y\in\mathcal{I}$ (that may depend on that given $x$) such that...

If you want to say "for every single $x$ in $\mathcal{I}$, one "fixed" $y$ works fine," then you would write $\exists y\in\mathcal{I},\,\forall x\in\mathcal{I},\,..$ Note the difference between the two expressions.

Therefore, $\forall x\in\mathbb{Z},\,\exists y\in\mathbb{Z},\,x+y=1$ is true, as we can choose $y = 1-x$. On the other hand, $\exists y\in\mathbb{Z},\,\forall x\in\mathbb{Z},\,x+y = 1$ is false: Assume such a $y$ exists. Then, pick $x = -y+123\in\mathbb{Z}$, then $x+y = 123$.

Edit: To make things more readable, I sometimes write, for example, $\forall x\in\mathcal{I},\,\exists y(x)\in\mathcal{I},\,...$ instead of $\forall x\in\mathcal{I},\,\exists y\in\mathcal{I},\,...$ in order not to get confused. Perhaps this may also work for you.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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