Mixing
Prove Triangle Inequality for Two Special Cases
$begingroup$
I am reading "Understanding Analysis" - it is a great book. However, I lack a solutions manual and some of these answers are difficult to find. It has been a while since I've written proofs and I'm wondering if these proofs are sufficient for the problem, or I can do better:
The problem states:
Use the triangle inequality to establish the inequalities:
(a) $|a - b| le |a| + |b|$
(b) $||a| - |b|| le |a - b|$
Starting with (a):
Note that $|a - b| = |a + (-b)|$
so $|a + (-b)| le |a| + |-b|$ by the definition of the Triangle Inequality.
So $|a + (-b)| le a + b$ by applying the rules of the absolute value.
By definition $|a + b| le a + b$
which surely means $|a + b| le |a| + |b|$ by the rules of the absolute value.
and since we have shown $|a + (-b)| le a + b$, $|a + b| le a + b$ and, $|a + b| le |a| + |b|$
Then it must the be case where $|a - b| le |a| + |b|$
For (b) it seems we can proceed directly:
$||a| - |b|| = |a - b|$ by the definition of the absolute value.
Then it must be the case that $|a - b| le |a - b|$
and so $||a| - |b|| le |a - b|$
I hope I did these correctly. It would be very helpful if I could be provided the correct answer if I am wrong - and methods to write cleaner proofs. Thank you!
real-analysis proof-verification
asked Jan 24 at 0:26
CL40CL40
2136
$endgroup$
add a comment |
$begingroup$
I am reading "Understanding Analysis" - it is a great book. However, I lack a solutions manual and some of these answers are difficult to find. It has been a while since I've written proofs and I'm wondering if these proofs are sufficient for the problem, or I can do better:
The problem states:
Use the triangle inequality to establish the inequalities:
(a) $|a - b| le |a| + |b|$
(b) $||a| - |b|| le |a - b|$
Starting with (a):
Note that $|a - b| = |a + (-b)|$
so $|a + (-b)| le |a| + |-b|$ by the definition of the Triangle Inequality.
So $|a + (-b)| le a + b$ by applying the rules of the absolute value.
By definition $|a + b| le a + b$
which surely means $|a + b| le |a| + |b|$ by the rules of the absolute value.
and since we have shown $|a + (-b)| le a + b$, $|a + b| le a + b$ and, $|a + b| le |a| + |b|$
Then it must the be case where $|a - b| le |a| + |b|$
For (b) it seems we can proceed directly:
$||a| - |b|| = |a - b|$ by the definition of the absolute value.
Then it must be the case that $|a - b| le |a - b|$
and so $||a| - |b|| le |a - b|$
I hope I did these correctly. It would be very helpful if I could be provided the correct answer if I am wrong - and methods to write cleaner proofs. Thank you!
real-analysis proof-verification
asked Jan 24 at 0:26
CL40CL40
2136
$endgroup$
add a comment |
$begingroup$
I am reading "Understanding Analysis" - it is a great book. However, I lack a solutions manual and some of these answers are difficult to find. It has been a while since I've written proofs and I'm wondering if these proofs are sufficient for the problem, or I can do better:
The problem states:
Use the triangle inequality to establish the inequalities:
(a) $|a - b| le |a| + |b|$
(b) $||a| - |b|| le |a - b|$
Starting with (a):
Note that $|a - b| = |a + (-b)|$
so $|a + (-b)| le |a| + |-b|$ by the definition of the Triangle Inequality.
So $|a + (-b)| le a + b$ by applying the rules of the absolute value.
By definition $|a + b| le a + b$
which surely means $|a + b| le |a| + |b|$ by the rules of the absolute value.
and since we have shown $|a + (-b)| le a + b$, $|a + b| le a + b$ and, $|a + b| le |a| + |b|$
Then it must the be case where $|a - b| le |a| + |b|$
For (b) it seems we can proceed directly:
$||a| - |b|| = |a - b|$ by the definition of the absolute value.
Then it must be the case that $|a - b| le |a - b|$
and so $||a| - |b|| le |a - b|$
I hope I did these correctly. It would be very helpful if I could be provided the correct answer if I am wrong - and methods to write cleaner proofs. Thank you!
real-analysis proof-verification
asked Jan 24 at 0:26
CL40CL40
2136
$endgroup$
I am reading "Understanding Analysis" - it is a great book. However, I lack a solutions manual and some of these answers are difficult to find. It has been a while since I've written proofs and I'm wondering if these proofs are sufficient for the problem, or I can do better:
The problem states:
Use the triangle inequality to establish the inequalities:
(a) $|a - b| le |a| + |b|$
(b) $||a| - |b|| le |a - b|$
Starting with (a):
Note that $|a - b| = |a + (-b)|$
so $|a + (-b)| le |a| + |-b|$ by the definition of the Triangle Inequality.
So $|a + (-b)| le a + b$ by applying the rules of the absolute value.
By definition $|a + b| le a + b$
which surely means $|a + b| le |a| + |b|$ by the rules of the absolute value.
and since we have shown $|a + (-b)| le a + b$, $|a + b| le a + b$ and, $|a + b| le |a| + |b|$
Then it must the be case where $|a - b| le |a| + |b|$
For (b) it seems we can proceed directly:
$||a| - |b|| = |a - b|$ by the definition of the absolute value.
Then it must be the case that $|a - b| le |a - b|$
and so $||a| - |b|| le |a - b|$
I hope I did these correctly. It would be very helpful if I could be provided the correct answer if I am wrong - and methods to write cleaner proofs. Thank you!
real-analysis proof-verification
real-analysis proof-verification
asked Jan 24 at 0:26
CL40CL40
2136
asked Jan 24 at 0:26
CL40CL40
2136
asked Jan 24 at 0:26
CL40CL40
2136
asked Jan 24 at 0:26
CL40CL40
2136
asked Jan 24 at 0:26
CL40CL40
2136
2136
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
(a) Prove $|a - b| le |a| + |b|$
We know that $|x|+|y| ge |x+y|$
$$text{Let $x=a$ and let $y=-b$. Substituting we get}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |-b| &ge |a + (-b)| \
|a| + |b| &ge |a-b| \
|a-b| &le |a| + |b|
end{align}
(b) Prove $||a| - |b|| le |a - b|$
$$text{Let $x=a$ and $y=b-a$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |b-a| &ge |b| \
|a| + |a-b| &ge |b| \
|b| &le |a| + |a-b| \
-|a-b| &le |a| - |b| tag{1.}
end{align}
$$text{Next, let $x=b$ and $y=a-b$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|b| + |a-b| &ge |a| \
|a| &le |b| + |a-b| \
|a| - |b| &le |a-b| tag{2.}
end{align}
$$text{From $(1.)$ and $(2.)$, we get}$$
begin{align}
-|a-b| le |a| - |b| le |a-b| \
||a| - |b|| le |a - b|
end{align}
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
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$begingroup$
If you have the already proven the standard form of the triangle inequality.
$|a+b| le |a|+|b|$
then you can say.
$|a-b| = |a+(-b)| le |a|+|-b| = |a|+|b|$
But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| le a+b$" when, in fact, $|a+b| ge a+b$
As for the second one, I am not following your logic. Here is what I might say.
$||a|-|b|| = ||(a-b)+b| - |b|| le ||a-b| + |b| - |b|| = |a-b|$
If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:
$age 0$ and $bge 0, a<0$ and $bge 0, age 0$ and $b< 0, a< 0$ and $b< 0.$
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$ lvert a+b rvert leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ lvert a+b rvert geq a+b$.
You need to use two facts for proving the first inequality:
(1) $a+(-b)=a-b$
(2) $ lvert -b rvert= lvert b rvert$
For (b), $lvert lvert a rvert -lvert brvert rvert=lvert a-brvert$ is untrue. A counter examples is a=1 and b=-1.
A correct reasoning would be noting $lvert a-b rvert=lvert b-a rvert$.
And $lvert a-b rvert +lvert b rvert geq lvert (a-b)+b rvert=lvert a rvert$ by the trinagle inequality.
Similarly $lvert b-a rvert +lvert a rvert geq lvert (b-a)+a rvert=lvert b rvert$
Therefore we have $lvert a-b rvert geq max(lvert a rvert-lvert b rvert,lvert b rvert-lvert a rvert)=lvert lvert a rvert-lvert b rvert rvert$ by the definition of the absolute value.
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(a) Prove $|a - b| le |a| + |b|$
We know that $|x|+|y| ge |x+y|$
$$text{Let $x=a$ and let $y=-b$. Substituting we get}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |-b| &ge |a + (-b)| \
|a| + |b| &ge |a-b| \
|a-b| &le |a| + |b|
end{align}
(b) Prove $||a| - |b|| le |a - b|$
$$text{Let $x=a$ and $y=b-a$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |b-a| &ge |b| \
|a| + |a-b| &ge |b| \
|b| &le |a| + |a-b| \
-|a-b| &le |a| - |b| tag{1.}
end{align}
$$text{Next, let $x=b$ and $y=a-b$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|b| + |a-b| &ge |a| \
|a| &le |b| + |a-b| \
|a| - |b| &le |a-b| tag{2.}
end{align}
$$text{From $(1.)$ and $(2.)$, we get}$$
begin{align}
-|a-b| le |a| - |b| le |a-b| \
||a| - |b|| le |a - b|
end{align}
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
$endgroup$
add a comment |
$begingroup$
(a) Prove $|a - b| le |a| + |b|$
We know that $|x|+|y| ge |x+y|$
$$text{Let $x=a$ and let $y=-b$. Substituting we get}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |-b| &ge |a + (-b)| \
|a| + |b| &ge |a-b| \
|a-b| &le |a| + |b|
end{align}
(b) Prove $||a| - |b|| le |a - b|$
$$text{Let $x=a$ and $y=b-a$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |b-a| &ge |b| \
|a| + |a-b| &ge |b| \
|b| &le |a| + |a-b| \
-|a-b| &le |a| - |b| tag{1.}
end{align}
$$text{Next, let $x=b$ and $y=a-b$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|b| + |a-b| &ge |a| \
|a| &le |b| + |a-b| \
|a| - |b| &le |a-b| tag{2.}
end{align}
$$text{From $(1.)$ and $(2.)$, we get}$$
begin{align}
-|a-b| le |a| - |b| le |a-b| \
||a| - |b|| le |a - b|
end{align}
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
$endgroup$
add a comment |
$begingroup$
(a) Prove $|a - b| le |a| + |b|$
We know that $|x|+|y| ge |x+y|$
$$text{Let $x=a$ and let $y=-b$. Substituting we get}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |-b| &ge |a + (-b)| \
|a| + |b| &ge |a-b| \
|a-b| &le |a| + |b|
end{align}
(b) Prove $||a| - |b|| le |a - b|$
$$text{Let $x=a$ and $y=b-a$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |b-a| &ge |b| \
|a| + |a-b| &ge |b| \
|b| &le |a| + |a-b| \
-|a-b| &le |a| - |b| tag{1.}
end{align}
$$text{Next, let $x=b$ and $y=a-b$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|b| + |a-b| &ge |a| \
|a| &le |b| + |a-b| \
|a| - |b| &le |a-b| tag{2.}
end{align}
$$text{From $(1.)$ and $(2.)$, we get}$$
begin{align}
-|a-b| le |a| - |b| le |a-b| \
||a| - |b|| le |a - b|
end{align}
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
$endgroup$
(a) Prove $|a - b| le |a| + |b|$
We know that $|x|+|y| ge |x+y|$
$$text{Let $x=a$ and let $y=-b$. Substituting we get}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |-b| &ge |a + (-b)| \
|a| + |b| &ge |a-b| \
|a-b| &le |a| + |b|
end{align}
(b) Prove $||a| - |b|| le |a - b|$
$$text{Let $x=a$ and $y=b-a$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|a| + |b-a| &ge |b| \
|a| + |a-b| &ge |b| \
|b| &le |a| + |a-b| \
-|a-b| &le |a| - |b| tag{1.}
end{align}
$$text{Next, let $x=b$ and $y=a-b$.}$$
begin{align}
|x|+|y| &ge |x+y| \
|b| + |a-b| &ge |a| \
|a| &le |b| + |a-b| \
|a| - |b| &le |a-b| tag{2.}
end{align}
$$text{From $(1.)$ and $(2.)$, we get}$$
begin{align}
-|a-b| le |a| - |b| le |a-b| \
||a| - |b|| le |a - b|
end{align}
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
answered Jan 24 at 1:22
steven gregorysteven gregory
18.3k32258
18.3k32258
add a comment |
add a comment |
$begingroup$
If you have the already proven the standard form of the triangle inequality.
$|a+b| le |a|+|b|$
then you can say.
$|a-b| = |a+(-b)| le |a|+|-b| = |a|+|b|$
But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| le a+b$" when, in fact, $|a+b| ge a+b$
As for the second one, I am not following your logic. Here is what I might say.
$||a|-|b|| = ||(a-b)+b| - |b|| le ||a-b| + |b| - |b|| = |a-b|$
If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:
$age 0$ and $bge 0, a<0$ and $bge 0, age 0$ and $b< 0, a< 0$ and $b< 0.$
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
If you have the already proven the standard form of the triangle inequality.
$|a+b| le |a|+|b|$
then you can say.
$|a-b| = |a+(-b)| le |a|+|-b| = |a|+|b|$
But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| le a+b$" when, in fact, $|a+b| ge a+b$
As for the second one, I am not following your logic. Here is what I might say.
$||a|-|b|| = ||(a-b)+b| - |b|| le ||a-b| + |b| - |b|| = |a-b|$
If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:
$age 0$ and $bge 0, a<0$ and $bge 0, age 0$ and $b< 0, a< 0$ and $b< 0.$
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
If you have the already proven the standard form of the triangle inequality.
$|a+b| le |a|+|b|$
then you can say.
$|a-b| = |a+(-b)| le |a|+|-b| = |a|+|b|$
But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| le a+b$" when, in fact, $|a+b| ge a+b$
As for the second one, I am not following your logic. Here is what I might say.
$||a|-|b|| = ||(a-b)+b| - |b|| le ||a-b| + |b| - |b|| = |a-b|$
If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:
$age 0$ and $bge 0, a<0$ and $bge 0, age 0$ and $b< 0, a< 0$ and $b< 0.$
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
$endgroup$
If you have the already proven the standard form of the triangle inequality.
$|a+b| le |a|+|b|$
then you can say.
$|a-b| = |a+(-b)| le |a|+|-b| = |a|+|b|$
But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| le a+b$" when, in fact, $|a+b| ge a+b$
As for the second one, I am not following your logic. Here is what I might say.
$||a|-|b|| = ||(a-b)+b| - |b|| le ||a-b| + |b| - |b|| = |a-b|$
If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:
$age 0$ and $bge 0, a<0$ and $bge 0, age 0$ and $b< 0, a< 0$ and $b< 0.$
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
answered Jan 24 at 1:02
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
$begingroup$
$ lvert a+b rvert leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ lvert a+b rvert geq a+b$.
You need to use two facts for proving the first inequality:
(1) $a+(-b)=a-b$
(2) $ lvert -b rvert= lvert b rvert$
For (b), $lvert lvert a rvert -lvert brvert rvert=lvert a-brvert$ is untrue. A counter examples is a=1 and b=-1.
A correct reasoning would be noting $lvert a-b rvert=lvert b-a rvert$.
And $lvert a-b rvert +lvert b rvert geq lvert (a-b)+b rvert=lvert a rvert$ by the trinagle inequality.
Similarly $lvert b-a rvert +lvert a rvert geq lvert (b-a)+a rvert=lvert b rvert$
Therefore we have $lvert a-b rvert geq max(lvert a rvert-lvert b rvert,lvert b rvert-lvert a rvert)=lvert lvert a rvert-lvert b rvert rvert$ by the definition of the absolute value.
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
$endgroup$
add a comment |
$begingroup$
$ lvert a+b rvert leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ lvert a+b rvert geq a+b$.
You need to use two facts for proving the first inequality:
(1) $a+(-b)=a-b$
(2) $ lvert -b rvert= lvert b rvert$
For (b), $lvert lvert a rvert -lvert brvert rvert=lvert a-brvert$ is untrue. A counter examples is a=1 and b=-1.
A correct reasoning would be noting $lvert a-b rvert=lvert b-a rvert$.
And $lvert a-b rvert +lvert b rvert geq lvert (a-b)+b rvert=lvert a rvert$ by the trinagle inequality.
Similarly $lvert b-a rvert +lvert a rvert geq lvert (b-a)+a rvert=lvert b rvert$
Therefore we have $lvert a-b rvert geq max(lvert a rvert-lvert b rvert,lvert b rvert-lvert a rvert)=lvert lvert a rvert-lvert b rvert rvert$ by the definition of the absolute value.
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
$endgroup$
add a comment |
$begingroup$
$ lvert a+b rvert leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ lvert a+b rvert geq a+b$.
You need to use two facts for proving the first inequality:
(1) $a+(-b)=a-b$
(2) $ lvert -b rvert= lvert b rvert$
For (b), $lvert lvert a rvert -lvert brvert rvert=lvert a-brvert$ is untrue. A counter examples is a=1 and b=-1.
A correct reasoning would be noting $lvert a-b rvert=lvert b-a rvert$.
And $lvert a-b rvert +lvert b rvert geq lvert (a-b)+b rvert=lvert a rvert$ by the trinagle inequality.
Similarly $lvert b-a rvert +lvert a rvert geq lvert (b-a)+a rvert=lvert b rvert$
Therefore we have $lvert a-b rvert geq max(lvert a rvert-lvert b rvert,lvert b rvert-lvert a rvert)=lvert lvert a rvert-lvert b rvert rvert$ by the definition of the absolute value.
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$ lvert a+b rvert leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ lvert a+b rvert geq a+b$.
You need to use two facts for proving the first inequality:
(1) $a+(-b)=a-b$
(2) $ lvert -b rvert= lvert b rvert$
For (b), $lvert lvert a rvert -lvert brvert rvert=lvert a-brvert$ is untrue. A counter examples is a=1 and b=-1.
A correct reasoning would be noting $lvert a-b rvert=lvert b-a rvert$.
And $lvert a-b rvert +lvert b rvert geq lvert (a-b)+b rvert=lvert a rvert$ by the trinagle inequality.
Similarly $lvert b-a rvert +lvert a rvert geq lvert (b-a)+a rvert=lvert b rvert$
Therefore we have $lvert a-b rvert geq max(lvert a rvert-lvert b rvert,lvert b rvert-lvert a rvert)=lvert lvert a rvert-lvert b rvert rvert$ by the definition of the absolute value.
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
answered Jan 24 at 1:02
NotoriousJuanGNotoriousJuanG
843
843
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