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Please help me out of this I am easily confuse by this kind of question.

Determine whether $∀x∈ℝ,∃y∈ℝ$ such that $x+y=0$ is true.

Logical thinking I know that it is true because for all $x$, I can choose a corresponding $y$ that will satisfy $x+y=0$. But should it be $x = -y$ or $ y =-x$?

Determine whether $∃x∈ℝ,∀y∈ℝ$ such that $x+y=0$ is true. I will say that it is false because there only one $y$ that can make $x+y=0.$

Thanks!

user292965
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For the first part, the equations $x=-y$ and $y=-x$ are actually equivalent, but what you need to do is:

Given a certain value of $x$, you need to find the appropriate value of $y$ such that $x+y=0$.

So, your result must be something that defines what value of $y$ you are taking. For example, if $x=4$, what value of $y$ do you take?

Second part: Your intuition is correct, now you need to put that into a rigorous proof. That is, you are disproving the statement:

$\exists x\in\mathbb R:\forall y\in\mathbb R: x+y=0$

Which means you must prove the negation of this statement, which is:

$\forall x\in\mathbb R: \exists y\in\mathbb R: x+y\neq 0$

To prove this statement, you must pick an arbitrary $x$ and find some $y$ for which $x+y$ is not equal to $0$.

5xum
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  • Hi 5xum,

    Thanks! Does it mean that for the first version it doesn't matter if I use x=−yx=−y or y=−x. Either answer will make it true?

    For the second question determine whether ∃x∈R,∀y∈R∃x∈ℝ,∀y∈ℝ such that x+y=0x+y=0 is true. Can I say something like x + y = x + (y+2), by contradiction 0≠2 so the statement is false?

     – user292965 Feb 23 '16 at 10:01
  • @user292965 No, for the first part, you need to define the value of $y$. I don't know what $y$ is, and you need to tell me. For example, what value of $y$ should I take if $x=4$? – 5xum Feb 23 '16 at 10:03
  • @user292965 For the second part, you cannot say that because that makes little sense and is not a mathematical proof. I can't make heads and tails of it. I don't see where the proof starts and where it ends. I don't see what your premise is and what you are proving. The "proof" is a mess of statements with no coherence. As I said: pick an arbitrary value of $x$, then set some value of $y$ for which $x+y=0$ is not true. – 5xum Feb 23 '16 at 10:05
  • from the question I think if u choose x = 4, I can take y =-4 and it will make it 0. Should it be done this way? – user292965 Feb 23 '16 at 10:05
  • @user292965 You are on the right track. You said "I can take $y=-4$". See? you detetermined what value of $y$ to take. Now, how about in general? If I give you an arbitrary $x$? What value of $y$ will you choose? – 5xum Feb 23 '16 at 10:06
  • I would choose y to be either the negative version of x or positive value of x. Dependings on whether x is positive or negative. – user292965 Feb 23 '16 at 10:10
  • @user292965 But in both cases, that would be $-x$, wouldn't it? See, if $x=-4$, then $-x=-(-4)=4$. – 5xum Feb 23 '16 at 10:12
  • does it mean that y = -x is correct? – user292965 Feb 23 '16 at 10:15
  • @user292965 Well, if it's correct, then you can prove it. Is it true that no matter what $x$ I choose, if you pick $y=-x$, then $x+y=0$? If this is true, then the choice $y=-x$ is correct. – 5xum Feb 23 '16 at 10:16
  • I understand it thanks! For question 2 when I disprove a statement and if I thurn out to be true does it means that the original statement is false? – user292965 Feb 23 '16 at 10:33
  • @user292965 precisely. To disprove a statement, you need to prove its negation. – 5xum Feb 23 '16 at 10:35
  • Thanks! Similarly I can also use the negation to prove the first question right? – user292965 Feb 23 '16 at 10:48
  • @user292965 Only if you want to disprove it. I wouldn't recommend disproving it, since you already proved it... – 5xum Feb 23 '16 at 10:50
  • Okay I will remember it. I am having problem trying to interpret the negation for the second question. After I negate "∃x∈R,∀y∈R such that x+y=0" to "∀x∈R,∃y∈R such that x+y≠0". I can read it as for all x belonging to the real number domain, there exists one y belonging to the real number domain such that x+y≠0. let say I take x as 4 and choose y as 5. Therefore the negated's statement is true and the original statement is false. Is it correct? – user292965 Feb 23 '16 at 11:22
  • @user292965 That's not good enough. You have a statement that starts with $\forall x$, you have to prove it for an arbitrary value of $x$. You cannot just set $x=4$. – 5xum Feb 23 '16 at 11:40
  • Do you mean I have to test all x value and it will satisfy the statement? If I test all the value including x = -4 when y = 4 will the statement become false? – user292965 Feb 23 '16 at 12:18
  • @user292965 No, I mean that you need to first say "let $x$ be arbitrary". Then, you find some $y$ (which can depend on the $x$ chosen) for which $x+y\neq 0$ is true. – 5xum Feb 23 '16 at 12:21
  • @user292965 Also, and I mean no offense, but you need to talk to somebody in person about this. Talk to your professor, or your school mates, or somebody. But you obviously have a very poor understanding of the process of mathematical proofs, and this site is no magic wand. – 5xum Feb 23 '16 at 12:22
  • You're nearly on $100k$ keep it up! +1 :D – Mr Pie Mar 12 '20 at 02:56