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 have the following two questions.

  1. For all real numbers x, there is a real number y such that $2x+y=7$ would this be true or false?

I think true because if you put


$2(8)+y=14$ there will always be a specific y that will make it work is this logic correct.

  1. There is a real numbers x that for all real number y, $2x+y=7$ will be true.

would this be false because if you say $x=6$

then you get


only if $y=2$ would it work but it would not work for every y.

share|cite|improve this question
Your answers are correct, but your justification for 1. isn't enough. Consider this: $$\forall x\in \Bbb R\exists y\in \Bbb R(2x+y=7)\iff \forall x\in \Bbb R\exists y\in \Bbb R(y=7-2x)$$ – Git Gud Nov 2 '13 at 19:27
Hmm so if you have $y=7-2x$ y must be a real number also x so my justification would be involve that they must equal. – Fernando Martinez Nov 2 '13 at 19:33
Do it the standard way: let $x$ be an arbitrary real number. You wish to prove that $\exists y\in \Bbb R(2x+y=7)$. Can you find a real number $y$ that works? – Git Gud Nov 2 '13 at 19:34
Yes if x is the arbitrary real number in $2x+y=7$ then y will be specifically $y=7-2x$ as you wrote on the first comment – Fernando Martinez Nov 2 '13 at 19:36
Exactly.${{{}}}$ – Git Gud Nov 2 '13 at 19:37
up vote 3 down vote accepted

Yes, indeed, you are correct in your assessment of the truth or falsity of each statement.

In the first, we can see this as allowing $y$ to depend on $x$. So for any given $x$, we can find some $y$, and in particular, we can simply choose $y = 7 - 2x$ which will guarantee the equalition holds.

In the second case, $y$ cannot depend on any given $x$. For the statement to be true, we need to consider the existence of a particular $y$ such that for every $x$, regardless of what $x$ may be, the equality holds. Since $x$ can vary, but $y$ can not vary accordingly, the statement is clearly false.

These two statements help demonstrate just how crucial the order of quantifiers and quantified variables can be: in the first, we have a true statement, and in the second, a false statement, and the only difference between them is the placement of $\exists y \cdots$.

share|cite|improve this answer

Yes, you're on the right track. Both statements can be more easily analyzed by rearranging to get

$$x = \frac{14 - y}{2}$$

This should allow you to fill in the gaps in your justification on (1) (i.e. not just stating a few examples in which it works).

share|cite|improve this answer
I don't see how solving for $x$ is helpful. Given $x$ we wish to find $y$. – Git Gud Nov 2 '13 at 19:31

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.