Mixing
Prove $sqrt{b} - sqrt{a} < sqrt{b-a}$
$begingroup$
Prove that if $0 < a < b$ then
$$sqrt{b} - sqrt{a} < sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2sqrt{ab} < b-a$
subtract $b$ from both sides $a-2sqrt{ab} < -a$
add $a$ to both sides $2a-2sqrt{ab} < 0$
than add $2 sqrt{ab}$ to both sides and get $2a < 2sqrt{ab}$
divide by $2$ and get $a < sqrt{ab}$
Thus we know $sqrt{ab}$ is bigger than $a$ so that $sqrt{a cdot a} < sqrt{ab}$ which means $sqrt{a} < sqrt{b}$.
Therefore together with the given $a < b$, we have $sqrt{b} - sqrt{a} < sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
algebra-precalculus inequality
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
$endgroup$
|
show 2 more comments
$begingroup$
Prove that if $0 < a < b$ then
$$sqrt{b} - sqrt{a} < sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2sqrt{ab} < b-a$
subtract $b$ from both sides $a-2sqrt{ab} < -a$
add $a$ to both sides $2a-2sqrt{ab} < 0$
than add $2 sqrt{ab}$ to both sides and get $2a < 2sqrt{ab}$
divide by $2$ and get $a < sqrt{ab}$
Thus we know $sqrt{ab}$ is bigger than $a$ so that $sqrt{a cdot a} < sqrt{ab}$ which means $sqrt{a} < sqrt{b}$.
Therefore together with the given $a < b$, we have $sqrt{b} - sqrt{a} < sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
algebra-precalculus inequality
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
$endgroup$
1
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
1
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
3
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
1
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58
|
show 2 more comments
$begingroup$
Prove that if $0 < a < b$ then
$$sqrt{b} - sqrt{a} < sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2sqrt{ab} < b-a$
subtract $b$ from both sides $a-2sqrt{ab} < -a$
add $a$ to both sides $2a-2sqrt{ab} < 0$
than add $2 sqrt{ab}$ to both sides and get $2a < 2sqrt{ab}$
divide by $2$ and get $a < sqrt{ab}$
Thus we know $sqrt{ab}$ is bigger than $a$ so that $sqrt{a cdot a} < sqrt{ab}$ which means $sqrt{a} < sqrt{b}$.
Therefore together with the given $a < b$, we have $sqrt{b} - sqrt{a} < sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
algebra-precalculus inequality
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
$endgroup$
Prove that if $0 < a < b$ then
$$sqrt{b} - sqrt{a} < sqrt{b-a}$$
This is what I have so far:
square both sides to get $a + b -2sqrt{ab} < b-a$
subtract $b$ from both sides $a-2sqrt{ab} < -a$
add $a$ to both sides $2a-2sqrt{ab} < 0$
than add $2 sqrt{ab}$ to both sides and get $2a < 2sqrt{ab}$
divide by $2$ and get $a < sqrt{ab}$
Thus we know $sqrt{ab}$ is bigger than $a$ so that $sqrt{a cdot a} < sqrt{ab}$ which means $sqrt{a} < sqrt{b}$.
Therefore together with the given $a < b$, we have $sqrt{b} - sqrt{a} < sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.
algebra-precalculus inequality
algebra-precalculus inequality
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
edited Feb 1 at 2:44
Lee David Chung Lin
4,47841242
4,47841242
asked Jan 31 at 23:38
user597188
1
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
1
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
3
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
1
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58
|
show 2 more comments
1
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
1
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
3
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
1
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58
1
1
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
1
1
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
3
3
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
1
1
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; tag 0$
then, note that
$b = a + (b - a) < a + 2sqrt a sqrt{b - a} + (b - a) = (sqrt a + sqrt{b - a})^2; tag 1$
thus,
$sqrt b < sqrt a + sqrt{b - a}, tag 2$
whence
$sqrt b - sqrt a < sqrt{b - a}. tag 3$
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2sqrt{a(b-a)}>0$, hence
$$2sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(sqrt{b-a}+sqrt{a})^2$, so
$$(sqrt{b-a}+sqrt{a})^2 > sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$sqrt{b-a}+sqrt{a} > sqrt{b},$$
which is your claim.
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
$endgroup$
add a comment |
$begingroup$
The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $acdot a < acdot b < bcdot b$ and so $a^2 < ab < b^2 $ and $a = sqrt a^2 < sqrt {ab} < sqrt b^2 = b$.
So $a < sqrt {ab}$
$2a < 2sqrt {ab}$
$ - 2sqrt{ab} +2a< 0$
$ -2sqrt{ab} +a< -a$
$b - 2sqrt{ab} + a < b - a$
$(sqrt b -sqrt a)^2 < b-a$
$|sqrt b - sqrt a| < sqrt{b-a}$.
But as $b > a > 0$ then $sqrt b > sqrt a$ and $|sqrt b - sqrt a| = sqrt b - sqrt a$.
So
$sqrt b - sqrt a < sqrt{b-a}$.
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
add a comment |
$begingroup$
You just need to note that every step you did is reversible. For the conclusion, from $alesqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=sqrt{a}$ and $y=sqrt{b}$. You need to show that
$$
y-x<sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
answered Jan 31 at 23:54
egregegreg
185k1486208
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; tag 0$
then, note that
$b = a + (b - a) < a + 2sqrt a sqrt{b - a} + (b - a) = (sqrt a + sqrt{b - a})^2; tag 1$
thus,
$sqrt b < sqrt a + sqrt{b - a}, tag 2$
whence
$sqrt b - sqrt a < sqrt{b - a}. tag 3$
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; tag 0$
then, note that
$b = a + (b - a) < a + 2sqrt a sqrt{b - a} + (b - a) = (sqrt a + sqrt{b - a})^2; tag 1$
thus,
$sqrt b < sqrt a + sqrt{b - a}, tag 2$
whence
$sqrt b - sqrt a < sqrt{b - a}. tag 3$
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; tag 0$
then, note that
$b = a + (b - a) < a + 2sqrt a sqrt{b - a} + (b - a) = (sqrt a + sqrt{b - a})^2; tag 1$
thus,
$sqrt b < sqrt a + sqrt{b - a}, tag 2$
whence
$sqrt b - sqrt a < sqrt{b - a}. tag 3$
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
$endgroup$
It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge. If the proof were more concise, it might be easier to decipher. Here's a suggestion:
We recall that
$0 < a < b; tag 0$
then, note that
$b = a + (b - a) < a + 2sqrt a sqrt{b - a} + (b - a) = (sqrt a + sqrt{b - a})^2; tag 1$
thus,
$sqrt b < sqrt a + sqrt{b - a}, tag 2$
whence
$sqrt b - sqrt a < sqrt{b - a}. tag 3$
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
answered Jan 31 at 23:57
Robert LewisRobert Lewis
48.9k23168
48.9k23168
add a comment |
add a comment |
$begingroup$
It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2sqrt{a(b-a)}>0$, hence
$$2sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(sqrt{b-a}+sqrt{a})^2$, so
$$(sqrt{b-a}+sqrt{a})^2 > sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$sqrt{b-a}+sqrt{a} > sqrt{b},$$
which is your claim.
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
$endgroup$
add a comment |
$begingroup$
It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2sqrt{a(b-a)}>0$, hence
$$2sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(sqrt{b-a}+sqrt{a})^2$, so
$$(sqrt{b-a}+sqrt{a})^2 > sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$sqrt{b-a}+sqrt{a} > sqrt{b},$$
which is your claim.
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
$endgroup$
add a comment |
$begingroup$
It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2sqrt{a(b-a)}>0$, hence
$$2sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(sqrt{b-a}+sqrt{a})^2$, so
$$(sqrt{b-a}+sqrt{a})^2 > sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$sqrt{b-a}+sqrt{a} > sqrt{b},$$
which is your claim.
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
$endgroup$
It looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2sqrt{a(b-a)}>0$, hence
$$2sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(sqrt{b-a}+sqrt{a})^2$, so
$$(sqrt{b-a}+sqrt{a})^2 > sqrt{b}^2$$
and since both sides are positive you can take the square root getting
$$sqrt{b-a}+sqrt{a} > sqrt{b},$$
which is your claim.
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
answered Jan 31 at 23:54
GibbsGibbs
5,4383927
5,4383927
add a comment |
add a comment |
$begingroup$
The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $acdot a < acdot b < bcdot b$ and so $a^2 < ab < b^2 $ and $a = sqrt a^2 < sqrt {ab} < sqrt b^2 = b$.
So $a < sqrt {ab}$
$2a < 2sqrt {ab}$
$ - 2sqrt{ab} +2a< 0$
$ -2sqrt{ab} +a< -a$
$b - 2sqrt{ab} + a < b - a$
$(sqrt b -sqrt a)^2 < b-a$
$|sqrt b - sqrt a| < sqrt{b-a}$.
But as $b > a > 0$ then $sqrt b > sqrt a$ and $|sqrt b - sqrt a| = sqrt b - sqrt a$.
So
$sqrt b - sqrt a < sqrt{b-a}$.
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
add a comment |
$begingroup$
The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $acdot a < acdot b < bcdot b$ and so $a^2 < ab < b^2 $ and $a = sqrt a^2 < sqrt {ab} < sqrt b^2 = b$.
So $a < sqrt {ab}$
$2a < 2sqrt {ab}$
$ - 2sqrt{ab} +2a< 0$
$ -2sqrt{ab} +a< -a$
$b - 2sqrt{ab} + a < b - a$
$(sqrt b -sqrt a)^2 < b-a$
$|sqrt b - sqrt a| < sqrt{b-a}$.
But as $b > a > 0$ then $sqrt b > sqrt a$ and $|sqrt b - sqrt a| = sqrt b - sqrt a$.
So
$sqrt b - sqrt a < sqrt{b-a}$.
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
add a comment |
$begingroup$
The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $acdot a < acdot b < bcdot b$ and so $a^2 < ab < b^2 $ and $a = sqrt a^2 < sqrt {ab} < sqrt b^2 = b$.
So $a < sqrt {ab}$
$2a < 2sqrt {ab}$
$ - 2sqrt{ab} +2a< 0$
$ -2sqrt{ab} +a< -a$
$b - 2sqrt{ab} + a < b - a$
$(sqrt b -sqrt a)^2 < b-a$
$|sqrt b - sqrt a| < sqrt{b-a}$.
But as $b > a > 0$ then $sqrt b > sqrt a$ and $|sqrt b - sqrt a| = sqrt b - sqrt a$.
So
$sqrt b - sqrt a < sqrt{b-a}$.
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
$endgroup$
The reasoning of your argument is basically correct but you can't argue from the conclusion to be the beginning without clarifying why.
(particularly you need to state what is known and why from what is being speculated and what required conditions are necessary for the speculation. You are stating things that you don't know are true as though they are true, then concluding things that would make them true as though those are results of them being true and finally stating a statement that is neither the result you are trying to prove nor your hypothesis without saying why it is relevant. [Your intent is that it is easily verified by the premise, and once verified we have already shown acceptance of it is sufficient for us to accept the conclusion.... Now if you are having trouble following the sequence of that argument... well, that's the exact same problem a reader of your proof will have.])
Use what you have done as a "rough draft" and write your argument in the correct "forward" direction.
$0 < a < b$ means that $acdot a < acdot b < bcdot b$ and so $a^2 < ab < b^2 $ and $a = sqrt a^2 < sqrt {ab} < sqrt b^2 = b$.
So $a < sqrt {ab}$
$2a < 2sqrt {ab}$
$ - 2sqrt{ab} +2a< 0$
$ -2sqrt{ab} +a< -a$
$b - 2sqrt{ab} + a < b - a$
$(sqrt b -sqrt a)^2 < b-a$
$|sqrt b - sqrt a| < sqrt{b-a}$.
But as $b > a > 0$ then $sqrt b > sqrt a$ and $|sqrt b - sqrt a| = sqrt b - sqrt a$.
So
$sqrt b - sqrt a < sqrt{b-a}$.
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
edited Feb 1 at 18:09
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
answered Feb 1 at 0:01
fleabloodfleablood
73.9k22891
73.9k22891
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
add a comment |
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
I made a few minor fixes to your latex. I'd also suggest replacing "Your argument is correct" with "Your argument has all the correct steps, but backwards and in the wrong order," because the argument isn't a correct proof of the claim; the OP might mistakenly infer that from your statement that it is correct.
$endgroup$
– John Hughes
Feb 1 at 12:55
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
$begingroup$
$2sqrt{ab} +a< -a$ should be $-2sqrt{ab} +a< -a$
$endgroup$
– Daniel Mathias
Feb 1 at 14:05
add a comment |
$begingroup$
You just need to note that every step you did is reversible. For the conclusion, from $alesqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=sqrt{a}$ and $y=sqrt{b}$. You need to show that
$$
y-x<sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
answered Jan 31 at 23:54
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
You just need to note that every step you did is reversible. For the conclusion, from $alesqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=sqrt{a}$ and $y=sqrt{b}$. You need to show that
$$
y-x<sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
answered Jan 31 at 23:54
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
You just need to note that every step you did is reversible. For the conclusion, from $alesqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=sqrt{a}$ and $y=sqrt{b}$. You need to show that
$$
y-x<sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
answered Jan 31 at 23:54
egregegreg
185k1486208
$endgroup$
You just need to note that every step you did is reversible. For the conclusion, from $alesqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=sqrt{a}$ and $y=sqrt{b}$. You need to show that
$$
y-x<sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
answered Jan 31 at 23:54
egregegreg
185k1486208
answered Jan 31 at 23:54
egregegreg
185k1486208
answered Jan 31 at 23:54
egregegreg
185k1486208
answered Jan 31 at 23:54
egregegreg
185k1486208
185k1486208
add a comment |
add a comment |
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1
$begingroup$
There is nothing wrong at all. What is there to fix?.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:41
1
$begingroup$
This is pretty good but you have to be careful when you square both sides since that doesn't always preserve inequalities. However if both sides are positive, it will.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:42
$begingroup$
im not sure he just said I had to fix It and show it both ways which is was confused about
$endgroup$
– user597188
Jan 31 at 23:45
3
$begingroup$
It's not clear how the flow of your argument is working. You are trying to prove $sqrt{a} - sqrt{b} < sqrt{a-b}$ so you can't start with that and go forward UNLESS you say something indicating, "this will be true if this is true and ... that will be true if this, and this IS true so". Without indicating what direction you are going in, I the reader, have no idea how to follow your argument.
$endgroup$
– fleablood
Jan 31 at 23:47
1
$begingroup$
No need to "show it both ways", you just have to show it one way - unfortunately, not the way you have given. In other words, the end of your argument should be the beginning, and the beginning should be the end.
$endgroup$
– David
Jan 31 at 23:58