Suppose that $x$ and $y$ are irrational, but $x + y$ is rational. Prove that $x -y$ is irrational. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rational + irrational = always irrational?Prove that the product of a rational and irrational number is irrationalRational + irrational = always irrational?Proof verification: Let $a$ be an irrational number and $r$ be a nonzero rational number. If $s$ is a rational number then $ar$ + $s$ is irrationalProve that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y < z < x$Prove that the square root of any irrational number is irrational.Given a rational number and an irrational number, both greater than 0, prove that the product between them is irrational.Prove that there is an irrational number and a rational number between any two distinct real numbersRational and irrational sequencesIs $sqrt2+sqrt3$ rational or irrational?Sum of two irrational numbers being rational or irrational
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Suppose that $x$ and $y$ are irrational, but $x + y$ is rational. Prove that $x -y$ is irrational.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rational + irrational = always irrational?Prove that the product of a rational and irrational number is irrationalRational + irrational = always irrational?Proof verification: Let $a$ be an irrational number and $r$ be a nonzero rational number. If $s$ is a rational number then $ar$ + $s$ is irrationalProve that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y < z < x$Prove that the square root of any irrational number is irrational.Given a rational number and an irrational number, both greater than 0, prove that the product between them is irrational.Prove that there is an irrational number and a rational number between any two distinct real numbersRational and irrational sequencesIs $sqrt2+sqrt3$ rational or irrational?Sum of two irrational numbers being rational or irrational
$begingroup$
i was wondering if someone could check my proof
$Q= a/b , c,d : a,c ∈ mathbb Z , b,d ∈ N>0$
$a/b =x+y$
$a/b -y=x$
proof by contradiction.
Let $x-y$ is rational
$c/d = x-y$
sub $a/b -y = x$ in for $x$
$c/d = (a/b -y) - y$
$a/b - c/d = 2y$
This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.
there it can be said $x-y$ is irrational
proof-verification irrational-numbers rational-numbers
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
i was wondering if someone could check my proof
$Q= a/b , c,d : a,c ∈ mathbb Z , b,d ∈ N>0$
$a/b =x+y$
$a/b -y=x$
proof by contradiction.
Let $x-y$ is rational
$c/d = x-y$
sub $a/b -y = x$ in for $x$
$c/d = (a/b -y) - y$
$a/b - c/d = 2y$
This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.
there it can be said $x-y$ is irrational
proof-verification irrational-numbers rational-numbers
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
1
$begingroup$
You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
$endgroup$
– lulu
Apr 8 at 12:18
$begingroup$
@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
$endgroup$
– Saucy O'Path
Apr 8 at 12:28
$begingroup$
What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
$endgroup$
– josh
Apr 8 at 12:33
|
show 1 more comment
$begingroup$
i was wondering if someone could check my proof
$Q= a/b , c,d : a,c ∈ mathbb Z , b,d ∈ N>0$
$a/b =x+y$
$a/b -y=x$
proof by contradiction.
Let $x-y$ is rational
$c/d = x-y$
sub $a/b -y = x$ in for $x$
$c/d = (a/b -y) - y$
$a/b - c/d = 2y$
This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.
there it can be said $x-y$ is irrational
proof-verification irrational-numbers rational-numbers
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
i was wondering if someone could check my proof
$Q= a/b , c,d : a,c ∈ mathbb Z , b,d ∈ N>0$
$a/b =x+y$
$a/b -y=x$
proof by contradiction.
Let $x-y$ is rational
$c/d = x-y$
sub $a/b -y = x$ in for $x$
$c/d = (a/b -y) - y$
$a/b - c/d = 2y$
This is a contradiction because $a/b$ and $c/d$ are not in their lowest terms.
there it can be said $x-y$ is irrational
proof-verification irrational-numbers rational-numbers
proof-verification irrational-numbers rational-numbers
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 9 at 11:37
josh
asked Apr 8 at 12:13
joshjosh
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:13
joshjosh
112
asked Apr 8 at 12:13
joshjosh
112
112
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
josh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
1
$begingroup$
You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
$endgroup$
– lulu
Apr 8 at 12:18
$begingroup$
@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
$endgroup$
– Saucy O'Path
Apr 8 at 12:28
$begingroup$
What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
$endgroup$
– josh
Apr 8 at 12:33
|
show 1 more comment
$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
1
$begingroup$
You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
$endgroup$
– lulu
Apr 8 at 12:18
$begingroup$
@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
$endgroup$
– Saucy O'Path
Apr 8 at 12:28
$begingroup$
What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
$endgroup$
– josh
Apr 8 at 12:33
$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
1
1
$begingroup$
You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
$endgroup$
– lulu
Apr 8 at 12:18
$begingroup$
You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
$endgroup$
– lulu
Apr 8 at 12:18
$begingroup$
@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
$endgroup$
– Saucy O'Path
Apr 8 at 12:28
$begingroup$
@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
$endgroup$
– Saucy O'Path
Apr 8 at 12:28
$begingroup$
What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
$endgroup$
– josh
Apr 8 at 12:33
$begingroup$
What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
$endgroup$
– josh
Apr 8 at 12:33
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Yes, your proof seems right.
I like the following writing.
If $x-yinmathbb Q$ so $x-y+x+y=2xinmathbb Q$, which says $xinmathbb Q$, which is a contradiction.
Id est, $x-ynotinmathbb Q$.
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
add a comment |
$begingroup$
Hint: $a,b inmathbb Qimplies dfracapm b2inmathbb Q$.
Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.
answered Apr 8 at 12:21
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
Sum of rational and irrational numbers is irrational. See here.
If $x,y$ are irrational and $x+y$ is rational, then:
$$x-y=(x+y)-2y textis irrational.$$
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, your proof seems right.
I like the following writing.
If $x-yinmathbb Q$ so $x-y+x+y=2xinmathbb Q$, which says $xinmathbb Q$, which is a contradiction.
Id est, $x-ynotinmathbb Q$.
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
add a comment |
$begingroup$
Yes, your proof seems right.
I like the following writing.
If $x-yinmathbb Q$ so $x-y+x+y=2xinmathbb Q$, which says $xinmathbb Q$, which is a contradiction.
Id est, $x-ynotinmathbb Q$.
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
add a comment |
$begingroup$
Yes, your proof seems right.
I like the following writing.
If $x-yinmathbb Q$ so $x-y+x+y=2xinmathbb Q$, which says $xinmathbb Q$, which is a contradiction.
Id est, $x-ynotinmathbb Q$.
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
Yes, your proof seems right.
I like the following writing.
If $x-yinmathbb Q$ so $x-y+x+y=2xinmathbb Q$, which says $xinmathbb Q$, which is a contradiction.
Id est, $x-ynotinmathbb Q$.
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
answered Apr 8 at 12:18
Michael RozenbergMichael Rozenberg
111k1896201
111k1896201
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
add a comment |
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
Can you further clarify what you mean, i don't quite understand.
$endgroup$
– josh
Apr 8 at 12:54
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
$begingroup$
The sum of two rational numbers it's a rational number. If x-y and x+y are rational numbers, so x-y+x+y it's a rational number, which says 2x is a rational number. I hope, now it's clear.
$endgroup$
– Michael Rozenberg
Apr 8 at 13:52
add a comment |
$begingroup$
Hint: $a,b inmathbb Qimplies dfracapm b2inmathbb Q$.
Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.
answered Apr 8 at 12:21
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
Hint: $a,b inmathbb Qimplies dfracapm b2inmathbb Q$.
Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.
answered Apr 8 at 12:21
useruser
6,56011031
$endgroup$
add a comment |
$begingroup$
Hint: $a,b inmathbb Qimplies dfracapm b2inmathbb Q$.
Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.
answered Apr 8 at 12:21
useruser
6,56011031
$endgroup$
Hint: $a,b inmathbb Qimplies dfracapm b2inmathbb Q$.
Apply this to $a=x+y$, $b=x-y$ to obtain a contradiction.
answered Apr 8 at 12:21
useruser
6,56011031
answered Apr 8 at 12:21
useruser
6,56011031
answered Apr 8 at 12:21
useruser
6,56011031
answered Apr 8 at 12:21
useruser
6,56011031
6,56011031
add a comment |
add a comment |
$begingroup$
Sum of rational and irrational numbers is irrational. See here.
If $x,y$ are irrational and $x+y$ is rational, then:
$$x-y=(x+y)-2y textis irrational.$$
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
$endgroup$
add a comment |
$begingroup$
Sum of rational and irrational numbers is irrational. See here.
If $x,y$ are irrational and $x+y$ is rational, then:
$$x-y=(x+y)-2y textis irrational.$$
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
$endgroup$
add a comment |
$begingroup$
Sum of rational and irrational numbers is irrational. See here.
If $x,y$ are irrational and $x+y$ is rational, then:
$$x-y=(x+y)-2y textis irrational.$$
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
$endgroup$
Sum of rational and irrational numbers is irrational. See here.
If $x,y$ are irrational and $x+y$ is rational, then:
$$x-y=(x+y)-2y textis irrational.$$
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
answered Apr 9 at 12:48
farruhotafarruhota
22.1k2942
22.1k2942
add a comment |
add a comment |
josh is a new contributor. Be nice, and check out our Code of Conduct.
josh is a new contributor. Be nice, and check out our Code of Conduct.
josh is a new contributor. Be nice, and check out our Code of Conduct.
josh is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What does $frac ab $ and $frac cd$ not being in their lowest terms (says who, by the way?) have to do with the concluion?
$endgroup$
– Saucy O'Path
Apr 8 at 12:15
$begingroup$
Definition of a rational number states that the denominator and numerator must be in their lowest forms
$endgroup$
– josh
Apr 8 at 12:18
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You were good up until the last step. Dividing by $2$ you showed that (under the assumption that $xpm yin mathbb Q$) $y$ is rational. That's all you need! Nothing at all to do with "lowest terms".
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– lulu
Apr 8 at 12:18
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@josh The fact that you did not write something in the form that definition demands doesn't mean that it is impossible to do so. It just means that you did not do it (allegedly, because you have no ground to assume that $operatornamegcd(a,b)ne1$ either).
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– Saucy O'Path
Apr 8 at 12:28
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What i am try to say, is since (a/b) - (c/d) have a common factor they are not in their lowest possible form and thus violate my assumption.i think i am a little lost on how to disprove my final line as to create a contradiction
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– josh
Apr 8 at 12:33