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Showing that $x$ is non-positive
Clash Royale CLAN TAG#URR8PPP
up vote
6
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favorite
I've just started to study analysis by myself and I'm having a hard time proving things. I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
$x, y ∈ R$
If for all $y>0$ we have $y≥x$ show that $0≥x$
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number because even the small numbers like $1/10^100000000$ are bigger than x. But how can I prove this in math? It seems pretty straightforward.
inequality real-numbers
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
asked Aug 12 at 17:16
70pr4k
586
add a comment |Â
up vote
6
down vote
favorite
I've just started to study analysis by myself and I'm having a hard time proving things. I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
$x, y ∈ R$
If for all $y>0$ we have $y≥x$ show that $0≥x$
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number because even the small numbers like $1/10^100000000$ are bigger than x. But how can I prove this in math? It seems pretty straightforward.
inequality real-numbers
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
asked Aug 12 at 17:16
70pr4k
586
Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Nope I haven't reached at that part yet.
– 70pr4k
Aug 12 at 17:21
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I've just started to study analysis by myself and I'm having a hard time proving things. I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
$x, y ∈ R$
If for all $y>0$ we have $y≥x$ show that $0≥x$
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number because even the small numbers like $1/10^100000000$ are bigger than x. But how can I prove this in math? It seems pretty straightforward.
inequality real-numbers
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
asked Aug 12 at 17:16
70pr4k
586
I've just started to study analysis by myself and I'm having a hard time proving things. I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
$x, y ∈ R$
If for all $y>0$ we have $y≥x$ show that $0≥x$
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number because even the small numbers like $1/10^100000000$ are bigger than x. But how can I prove this in math? It seems pretty straightforward.
inequality real-numbers
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
asked Aug 12 at 17:16
70pr4k
586
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
edited Aug 13 at 8:59
TheSimpliFire
10.5k62053
10.5k62053
asked Aug 12 at 17:16
70pr4k
586
asked Aug 12 at 17:16
70pr4k
586
asked Aug 12 at 17:16
70pr4k
586
586
Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Nope I haven't reached at that part yet.
– 70pr4k
Aug 12 at 17:21
add a comment |Â
Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Nope I haven't reached at that part yet.
– 70pr4k
Aug 12 at 17:21
Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Nope I haven't reached at that part yet.
– 70pr4k
Aug 12 at 17:21
Nope I haven't reached at that part yet.
– 70pr4k
Aug 12 at 17:21
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
13
down vote
accepted
If $x>0$, take $y=frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $ygeqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $xleqslant0$.
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
add a comment |Â
up vote
6
down vote
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...
The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
answered Aug 12 at 17:58
Allawonder
1,847516
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
add a comment |Â
up vote
2
down vote
Suppose
$x le 0 tag 1$
is false. Then
$x > 0; tag 2$
set
$y = dfrac12 x; tag 3$
then
$0 < y < x. tag 4$
We have just shown the existence of $y in Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$forall 0 < y in Bbb R, ; y ge x; tag 5$
therefore there is no such $x > 0$; thus (1) binds.
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
add a comment |Â
up vote
2
down vote
You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive.
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $xle 0$
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
If $x>0$, take $y=frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $ygeqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $xleqslant0$.
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
add a comment |Â
up vote
13
down vote
accepted
If $x>0$, take $y=frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $ygeqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $xleqslant0$.
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
If $x>0$, take $y=frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $ygeqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $xleqslant0$.
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
If $x>0$, take $y=frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $ygeqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $xleqslant0$.
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
answered Aug 12 at 17:19
José Carlos Santos
119k16101182
119k16101182
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
add a comment |Â
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
So when I assume x>0 to make a contradiction, I can just assign a value to y and I'm choosing $y=x/2$ to make it smaller than x so that It contradicts. And I can do that because we basically assumed that both x and y can be any positive real number at first. Did I understand it correctly?
– 70pr4k
Aug 12 at 17:33
1
1
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
@70pr4k Almost. The only problem is that I did not assume that $y>0$. I only assumed that $x>0$. Since $y=frac x2$, it is automatic that $y>0$.
– José Carlos Santos
Aug 12 at 17:35
Thank you so much!
– 70pr4k
Aug 12 at 17:37
Thank you so much!
– 70pr4k
Aug 12 at 17:37
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
@70pr4k I'm glad I could help.
– José Carlos Santos
Aug 12 at 17:38
add a comment |Â
up vote
6
down vote
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...
The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
answered Aug 12 at 17:58
Allawonder
1,847516
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
add a comment |Â
up vote
6
down vote
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...
The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
answered Aug 12 at 17:58
Allawonder
1,847516
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
add a comment |Â
up vote
6
down vote
up vote
6
down vote
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...
The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
answered Aug 12 at 17:58
Allawonder
1,847516
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...
The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
answered Aug 12 at 17:58
Allawonder
1,847516
edited Aug 12 at 18:13
answered Aug 12 at 17:58
Allawonder
1,847516
answered Aug 12 at 17:58
Allawonder
1,847516
answered Aug 12 at 17:58
Allawonder
1,847516
1,847516
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
add a comment |Â
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
1
1
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
Thank you so much! This really gave me some hope and morale!
– 70pr4k
Aug 12 at 18:03
add a comment |Â
up vote
2
down vote
Suppose
$x le 0 tag 1$
is false. Then
$x > 0; tag 2$
set
$y = dfrac12 x; tag 3$
then
$0 < y < x. tag 4$
We have just shown the existence of $y in Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$forall 0 < y in Bbb R, ; y ge x; tag 5$
therefore there is no such $x > 0$; thus (1) binds.
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
add a comment |Â
up vote
2
down vote
Suppose
$x le 0 tag 1$
is false. Then
$x > 0; tag 2$
set
$y = dfrac12 x; tag 3$
then
$0 < y < x. tag 4$
We have just shown the existence of $y in Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$forall 0 < y in Bbb R, ; y ge x; tag 5$
therefore there is no such $x > 0$; thus (1) binds.
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Suppose
$x le 0 tag 1$
is false. Then
$x > 0; tag 2$
set
$y = dfrac12 x; tag 3$
then
$0 < y < x. tag 4$
We have just shown the existence of $y in Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$forall 0 < y in Bbb R, ; y ge x; tag 5$
therefore there is no such $x > 0$; thus (1) binds.
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
Suppose
$x le 0 tag 1$
is false. Then
$x > 0; tag 2$
set
$y = dfrac12 x; tag 3$
then
$0 < y < x. tag 4$
We have just shown the existence of $y in Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$forall 0 < y in Bbb R, ; y ge x; tag 5$
therefore there is no such $x > 0$; thus (1) binds.
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
answered Aug 12 at 17:27
Robert Lewis
37.9k22357
37.9k22357
add a comment |Â
add a comment |Â
up vote
2
down vote
You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive.
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $xle 0$
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
add a comment |Â
up vote
2
down vote
You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive.
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $xle 0$
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive.
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $xle 0$
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive.
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $xle 0$
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
edited Aug 12 at 21:13
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
answered Aug 12 at 17:24
Mohammad Riazi-Kermani
30.3k41852
30.3k41852
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
add a comment |Â
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
This would work if the condition wasn't that for y>0 y is greater than or equal to x, rather than just greater than
– user399625
Aug 12 at 21:03
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
@user399625 Thanks for the comment, I have edited my answer to fix the error.
– Mohammad Riazi-Kermani
Aug 12 at 21:15
add a comment |Â
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Do you know about inf and sup?
– Benjamin Tighe
Aug 12 at 17:18
Nope I haven't reached at that part yet.
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Aug 12 at 17:21