# Discrete math logic question

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

$2(7)+y=14$

$2(8)+y=14$ 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

$2(6)+2=14$

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

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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

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 $\exists y \cdots$.

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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).

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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