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Apostol's mathematical analysis theorem 1.1
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Theorem 1.1. Given real numbers $a$ and $b$ such that $$a leq b + epsilon, text for every epsilon gt 0,text then a leq b tag 1$$
Proof. If $$b lt a,$$ then inequality $(1)$ is violated for $$epsilon = (a-b)/2 $$because
$$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a $$
I understand Theorem, but I didn't understand the proof part.
why $$epsilon = (a-b)/2$$ comes out, and what $$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a$$ stands for?
analysis
edited Jul 24 at 0:14
Bernard
110k635103
asked Jul 23 at 23:36
justin0526
62
add a comment |Â
up vote
1
down vote
favorite
Theorem 1.1. Given real numbers $a$ and $b$ such that $$a leq b + epsilon, text for every epsilon gt 0,text then a leq b tag 1$$
Proof. If $$b lt a,$$ then inequality $(1)$ is violated for $$epsilon = (a-b)/2 $$because
$$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a $$
I understand Theorem, but I didn't understand the proof part.
why $$epsilon = (a-b)/2$$ comes out, and what $$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a$$ stands for?
analysis
edited Jul 24 at 0:14
Bernard
110k635103
asked Jul 23 at 23:36
justin0526
62
1
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$,and so forth
– Omnomnomnom
Jul 23 at 23:54
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem 1.1. Given real numbers $a$ and $b$ such that $$a leq b + epsilon, text for every epsilon gt 0,text then a leq b tag 1$$
Proof. If $$b lt a,$$ then inequality $(1)$ is violated for $$epsilon = (a-b)/2 $$because
$$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a $$
I understand Theorem, but I didn't understand the proof part.
why $$epsilon = (a-b)/2$$ comes out, and what $$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a$$ stands for?
analysis
edited Jul 24 at 0:14
Bernard
110k635103
asked Jul 23 at 23:36
justin0526
62
Theorem 1.1. Given real numbers $a$ and $b$ such that $$a leq b + epsilon, text for every epsilon gt 0,text then a leq b tag 1$$
Proof. If $$b lt a,$$ then inequality $(1)$ is violated for $$epsilon = (a-b)/2 $$because
$$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a $$
I understand Theorem, but I didn't understand the proof part.
why $$epsilon = (a-b)/2$$ comes out, and what $$b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a$$ stands for?
analysis
edited Jul 24 at 0:14
Bernard
110k635103
asked Jul 23 at 23:36
justin0526
62
edited Jul 24 at 0:14
Bernard
110k635103
edited Jul 24 at 0:14
Bernard
110k635103
edited Jul 24 at 0:14
Bernard
110k635103
110k635103
asked Jul 23 at 23:36
justin0526
62
asked Jul 23 at 23:36
justin0526
62
asked Jul 23 at 23:36
justin0526
62
62
1
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$,and so forth
– Omnomnomnom
Jul 23 at 23:54
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57
add a comment |Â
1
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$,and so forth
– Omnomnomnom
Jul 23 at 23:54
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57
1
1
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:
real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$, and so forth– Omnomnomnom
Jul 23 at 23:54
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:
real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$, and so forth– Omnomnomnom
Jul 23 at 23:54
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
The line
$$
b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a
$$
stands for a proof of $b+epsilon < a$, done in several steps:
$$
b + epsilon = b + (a-b)/2,
\
b + (a-b)/2 = (a+b)/2
\
(a+b)/2 lt (a+a)/2
\
(a+a)/2 = a
$$
added
Why $epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $epsilon$. This particular choice for $epsilon$ was made because the calculation (see above) is simple for it.
answered Jul 23 at 23:55
GEdgar
58.4k264163
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The line
$$
b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a
$$
stands for a proof of $b+epsilon < a$, done in several steps:
$$
b + epsilon = b + (a-b)/2,
\
b + (a-b)/2 = (a+b)/2
\
(a+b)/2 lt (a+a)/2
\
(a+a)/2 = a
$$
added
Why $epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $epsilon$. This particular choice for $epsilon$ was made because the calculation (see above) is simple for it.
answered Jul 23 at 23:55
GEdgar
58.4k264163
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
add a comment |Â
up vote
1
down vote
The line
$$
b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a
$$
stands for a proof of $b+epsilon < a$, done in several steps:
$$
b + epsilon = b + (a-b)/2,
\
b + (a-b)/2 = (a+b)/2
\
(a+b)/2 lt (a+a)/2
\
(a+a)/2 = a
$$
added
Why $epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $epsilon$. This particular choice for $epsilon$ was made because the calculation (see above) is simple for it.
answered Jul 23 at 23:55
GEdgar
58.4k264163
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The line
$$
b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a
$$
stands for a proof of $b+epsilon < a$, done in several steps:
$$
b + epsilon = b + (a-b)/2,
\
b + (a-b)/2 = (a+b)/2
\
(a+b)/2 lt (a+a)/2
\
(a+a)/2 = a
$$
added
Why $epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $epsilon$. This particular choice for $epsilon$ was made because the calculation (see above) is simple for it.
answered Jul 23 at 23:55
GEdgar
58.4k264163
The line
$$
b + epsilon = b + (a-b)/2 = (a+b)/2 lt (a+a)/2 = a
$$
stands for a proof of $b+epsilon < a$, done in several steps:
$$
b + epsilon = b + (a-b)/2,
\
b + (a-b)/2 = (a+b)/2
\
(a+b)/2 lt (a+a)/2
\
(a+a)/2 = a
$$
added
Why $epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $epsilon$. This particular choice for $epsilon$ was made because the calculation (see above) is simple for it.
answered Jul 23 at 23:55
GEdgar
58.4k264163
edited Jul 24 at 0:20
answered Jul 23 at 23:55
GEdgar
58.4k264163
answered Jul 23 at 23:55
GEdgar
58.4k264163
answered Jul 23 at 23:55
GEdgar
58.4k264163
58.4k264163
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
add a comment |Â
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
It is reasonable, but why õ=(a−b)/2 comes? I think that confuses me
– justin0526
Jul 24 at 0:06
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
Picking the right constants to use in proofs like this is an art; it is not obvious at first but the more you practice the easier it gets.
– dbx
Jul 24 at 3:36
add a comment |Â
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1
It looks like you're having trouble with the formatting; see this tutorial. Note that you can write mathematics with "inline formatting" as follows:
real numbers $a$ and $b$ such that $a leq b$, for every $epsilon > 0$,and so forth– Omnomnomnom
Jul 23 at 23:54
@Omnomnomnom Yes, I have to learn it. thanks for the link
– justin0526
Jul 23 at 23:57