Dtjytyk

How to split IPA spelling into syllables

Why can't I get pgrep output right to variable on bash script?

How would a solely written language work mechanically

Asserting that Atheism and Theism are both faith based positions

Connection Between Knot Theory and Number Theory

Travelling in US for more than 90 days

Why is implicit conversion not ambiguous for non-primitive types?

What is the meaning of "You've never met a graph you didn't like?"

Are hand made posters acceptable in Academia?

Weird lines in Microsoft Word

Hashing password to increase entropy

Do native speakers use "ultima" and "proxima" frequently in spoken English?

What is the tangent at a sharp point on a curve?

Calculate Pi using Monte Carlo

How to preserve electronics (computers, ipads, phones) for hundreds of years?

Should a narrator ever describe things based on a character's view instead of facts?

Extract substring according to regexp with sed or grep

When is the exact date for EOL of Ubuntu 14.04 LTS?

categorizing a variable turns it from insignificant to significant

How do I lift the insulation blower into the attic?

What is the purpose of using a decision tree?

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

Why doesn't Gödel's incompleteness theorem apply to false statements?

Put the phone down / Put down the phone

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

Prove true in natural numbers (Peano Arithmetic)How to determine if a predicate statement is true or false while giving reasons?Proving whether a statement is true or false; regards predicate logic.Deciding if a statement is true or false for given setsHow to prove that the following predicate formula is valid…Logic determine whether a set is consistentPredicates and Quantifiers_Discrete MathWorking out if a predicate formula is true or false given a domain {0, 1}7 True or False Questions: predicate logic - are my answers & proofs correct?Find the argument form for the argument and determine whether it is valid

0

$begingroup$

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!

predicate-logic

share|cite|improve this question

edited Feb 23 '16 at 9:55

user292965

asked Feb 23 '16 at 9:50

user292965user292965

1218

$endgroup$

add a comment | 

0

$begingroup$

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!

predicate-logic

share|cite|improve this question

edited Feb 23 '16 at 9:55

user292965

asked Feb 23 '16 at 9:50

user292965user292965

1218

$endgroup$

add a comment | 

$begingroup$

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!

predicate-logic

share|cite|improve this question

edited Feb 23 '16 at 9:55

user292965

asked Feb 23 '16 at 9:50

user292965user292965

1218

$endgroup$

share|cite|improve this question

edited Feb 23 '16 at 9:55

user292965

asked Feb 23 '16 at 9:50

user292965user292965

1218

share|cite|improve this question

edited Feb 23 '16 at 9:55

user292965

asked Feb 23 '16 at 9:50

user292965user292965

1218

asked Feb 23 '16 at 9:50

user292965user292965

1218

asked Feb 23 '16 at 9:50

user292965user292965

1218

1 Answer
1

active

oldest

votes

0

$begingroup$

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:

$exists xinmathbb R:forall yinmathbb R: x+y=0$

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

$forall xinmathbb R: exists yinmathbb R: x+yneq 0$

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

share|cite|improve this answer

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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 $x=4$?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:06
  

  
  
  
  

 | 
show 13 more comments

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1668388%2fdetermine-whether-%25e2%2588%2580x%25e2%2588%2588%25e2%2584%259d-%25e2%2588%2583y%25e2%2588%2588%25e2%2584%259d-such-that-xy-0-%25e2%2588%2583x%25e2%2588%2588%25e2%2584%259d-%25e2%2588%2580y%25e2%2588%2588%25e2%2584%259d-such-that-xy-0%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

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:

$exists xinmathbb R:forall yinmathbb R: x+y=0$

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

$forall xinmathbb R: exists yinmathbb R: x+yneq 0$

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

share|cite|improve this answer

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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 $x=4$?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:06
  

  
  
  
  

 | 
show 13 more comments

0

$begingroup$

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:

$exists xinmathbb R:forall yinmathbb R: x+y=0$

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

$forall xinmathbb R: exists yinmathbb R: x+yneq 0$

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

share|cite|improve this answer

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:01
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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 $x=4$?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – user292965
  Feb 23 '16 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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?
  $endgroup$
  – 5xum
  Feb 23 '16 at 10:06
  

  
  
  
  

 | 
show 13 more comments

$begingroup$

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:

$exists xinmathbb R:forall yinmathbb R: x+y=0$

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

$forall xinmathbb R: exists yinmathbb R: x+yneq 0$

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

share|cite|improve this answer

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

$endgroup$

share|cite|improve this answer

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161

answered Feb 23 '16 at 9:57

5xum5xum

91.4k394161