Mixing
If $x>sqrt{xy}>y$, then $x>y>0$.
I am trying to prove the following:
If $x>sqrt{xy}>y$, then show that $x>y>0$.
My argument is as follows:
We only need to show $y>0$. Suppose $y<0$. Then, for $sqrt{xy}$ to be defined, we need $x<0$. Thus, $x<sqrt{xy}$, which is a contradiction. Hence, $y>0$.
Somehow, my proof doesn't sound "rigorous" enough to me. Is there any other method to prove the result?
algebra-precalculus inequality radicals
edited Nov 10 '18 at 11:53
NoChance
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
|
show 2 more comments
I am trying to prove the following:
If $x>sqrt{xy}>y$, then show that $x>y>0$.
My argument is as follows:
We only need to show $y>0$. Suppose $y<0$. Then, for $sqrt{xy}$ to be defined, we need $x<0$. Thus, $x<sqrt{xy}$, which is a contradiction. Hence, $y>0$.
Somehow, my proof doesn't sound "rigorous" enough to me. Is there any other method to prove the result?
algebra-precalculus inequality radicals
edited Nov 10 '18 at 11:53
NoChance
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
1
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
1
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
3
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
A final tip: To get the square root symbol to extend over an expression, enclose the expression with{}. For example,$sqrt{xy}$.
– Blue
Nov 10 '18 at 11:36
1
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55
|
show 2 more comments
I am trying to prove the following:
If $x>sqrt{xy}>y$, then show that $x>y>0$.
My argument is as follows:
We only need to show $y>0$. Suppose $y<0$. Then, for $sqrt{xy}$ to be defined, we need $x<0$. Thus, $x<sqrt{xy}$, which is a contradiction. Hence, $y>0$.
Somehow, my proof doesn't sound "rigorous" enough to me. Is there any other method to prove the result?
algebra-precalculus inequality radicals
edited Nov 10 '18 at 11:53
NoChance
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
I am trying to prove the following:
If $x>sqrt{xy}>y$, then show that $x>y>0$.
My argument is as follows:
We only need to show $y>0$. Suppose $y<0$. Then, for $sqrt{xy}$ to be defined, we need $x<0$. Thus, $x<sqrt{xy}$, which is a contradiction. Hence, $y>0$.
Somehow, my proof doesn't sound "rigorous" enough to me. Is there any other method to prove the result?
algebra-precalculus inequality radicals
algebra-precalculus inequality radicals
edited Nov 10 '18 at 11:53
NoChance
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
edited Nov 10 '18 at 11:53
NoChance
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
edited Nov 10 '18 at 11:53
NoChance
3,63121221
edited Nov 10 '18 at 11:53
NoChance
3,63121221
edited Nov 10 '18 at 11:53
NoChance
3,63121221
3,63121221
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
asked Nov 10 '18 at 11:14
PGuptaPGupta
1373
1373
1
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
1
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
3
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
A final tip: To get the square root symbol to extend over an expression, enclose the expression with{}. For example,$sqrt{xy}$.
– Blue
Nov 10 '18 at 11:36
1
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55
|
show 2 more comments
1
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
1
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
3
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
A final tip: To get the square root symbol to extend over an expression, enclose the expression with{}. For example,$sqrt{xy}$.
– Blue
Nov 10 '18 at 11:36
1
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55
1
1
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
1
1
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
3
3
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
A final tip: To get the square root symbol to extend over an expression, enclose the expression with
{}. For example, $sqrt{xy}$.– Blue
Nov 10 '18 at 11:36
A final tip: To get the square root symbol to extend over an expression, enclose the expression with
{}. For example, $sqrt{xy}$.– Blue
Nov 10 '18 at 11:36
1
1
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55
|
show 2 more comments
2 Answers
2
active
oldest
votes
$x$ is positive since $x>sqrt {text{something}}$ which means that $y$ is also positive since $xy$ is positive and $yne sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need
$$xy> 0 implies x>sqrt{xy}>0 quad land quad y> 0$$
answered Nov 10 '18 at 11:43
gimusigimusi
1
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
|
show 1 more comment
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$x$ is positive since $x>sqrt {text{something}}$ which means that $y$ is also positive since $xy$ is positive and $yne sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
$x$ is positive since $x>sqrt {text{something}}$ which means that $y$ is also positive since $xy$ is positive and $yne sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
$x$ is positive since $x>sqrt {text{something}}$ which means that $y$ is also positive since $xy$ is positive and $yne sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
$x$ is positive since $x>sqrt {text{something}}$ which means that $y$ is also positive since $xy$ is positive and $yne sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 21 '18 at 21:24
Mostafa AyazMostafa Ayaz
14.1k3937
14.1k3937
add a comment |
add a comment |
Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need
$$xy> 0 implies x>sqrt{xy}>0 quad land quad y> 0$$
answered Nov 10 '18 at 11:43
gimusigimusi
1
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
|
show 1 more comment
Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need
$$xy> 0 implies x>sqrt{xy}>0 quad land quad y> 0$$
answered Nov 10 '18 at 11:43
gimusigimusi
1
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
|
show 1 more comment
Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need
$$xy> 0 implies x>sqrt{xy}>0 quad land quad y> 0$$
answered Nov 10 '18 at 11:43
gimusigimusi
1
Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need
$$xy> 0 implies x>sqrt{xy}>0 quad land quad y> 0$$
answered Nov 10 '18 at 11:43
gimusigimusi
1
edited Nov 10 '18 at 11:55
answered Nov 10 '18 at 11:43
gimusigimusi
1
answered Nov 10 '18 at 11:43
gimusigimusi
1
answered Nov 10 '18 at 11:43
gimusigimusi
1
1
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
|
show 1 more comment
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
You are told that $x>y.$ You want to show that $y>0.$
– Will R
Nov 10 '18 at 11:45
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
@WillR Yes I've added some useless step! Thanks I revise that.
– gimusi
Nov 10 '18 at 11:49
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
It's still a bit weird. Why do you start squaring things?
– Will R
Nov 10 '18 at 11:50
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
@WillR Yes indeed I was referring to that part! The chain is $$xy>0 implies x>0 implies y>0$$
– gimusi
Nov 10 '18 at 11:52
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
Yes, that is the chain.
– Will R
Nov 10 '18 at 11:53
|
show 1 more comment
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1
I am fine with that
– Hagen von Eitzen
Nov 10 '18 at 11:19
1
There's a very minor oversight: You should address that $y=0$ is also problematic.
– Blue
Nov 10 '18 at 11:22
3
Your argument is good, but makes unnecessary use of contradiction. You can turn this into a direct approach like this: Square roots are non-negative, so $x>sqrt{xy} geq 0$, which implies $x > 0$. For $sqrt{xy}$ to be defined (and real), $xygeq 0$. Therefore, $y geq 0$. Since we cannot have $y=0$ (why?), we have $y>0$. $square$
– Blue
Nov 10 '18 at 11:33
A final tip: To get the square root symbol to extend over an expression, enclose the expression with
{}. For example,$sqrt{xy}$.– Blue
Nov 10 '18 at 11:36
1
@Blue Thanks for the LaTeX suggestion: I was mistakenly trying with round brackets and it wasn't working. Also thanks for the alternate argument: it makes sense.
– PGupta
Nov 14 '18 at 4:55