Proposition 2.2.12 Analysis I Tao
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$(f)$ $a<b$ $if$ $and$ $only$ $if$ $b=a+d$ $for$ $some$ $positive$ $number$ $d$
In my book $a<b$ is defined as $aleq b$ $and$ $aneq b$
and $aleq b$ is by def. $b=a+d$ $for$ $some$ $natural$ $number$ $d$
By using this and Peano axioms, how do I prove (f)?
real-analysis
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
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add a comment |
$begingroup$
$(f)$ $a<b$ $if$ $and$ $only$ $if$ $b=a+d$ $for$ $some$ $positive$ $number$ $d$
In my book $a<b$ is defined as $aleq b$ $and$ $aneq b$
and $aleq b$ is by def. $b=a+d$ $for$ $some$ $natural$ $number$ $d$
By using this and Peano axioms, how do I prove (f)?
real-analysis
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
$endgroup$
$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
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Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53
add a comment |
$begingroup$
$(f)$ $a<b$ $if$ $and$ $only$ $if$ $b=a+d$ $for$ $some$ $positive$ $number$ $d$
In my book $a<b$ is defined as $aleq b$ $and$ $aneq b$
and $aleq b$ is by def. $b=a+d$ $for$ $some$ $natural$ $number$ $d$
By using this and Peano axioms, how do I prove (f)?
real-analysis
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
$endgroup$
$(f)$ $a<b$ $if$ $and$ $only$ $if$ $b=a+d$ $for$ $some$ $positive$ $number$ $d$
In my book $a<b$ is defined as $aleq b$ $and$ $aneq b$
and $aleq b$ is by def. $b=a+d$ $for$ $some$ $natural$ $number$ $d$
By using this and Peano axioms, how do I prove (f)?
real-analysis
real-analysis
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
edited Dec 13 '18 at 18:04
Fruchtsaft
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
asked Dec 13 '18 at 17:23
FruchtsaftFruchtsaft
11
11
$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
$begingroup$
Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53
add a comment |
$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
$begingroup$
Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53
$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
$begingroup$
Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $mathbb{N}$ denote the set of natural numbers and $a, b in mathbb{N}$.
1) Let $a < b$. Then $b = a + d$ for some $d in mathbb{N}$ and $a ne b$. Hence $d = 0$ is impossible because $a + 0 = a$, that is, $b = a + d$ for some positive $d in mathbb{N}$.
2) Let $b = a + d$ for some positive $d in mathbb{N}$. Then certainly $a le b$. But $a = b$ is impossible because in that case $a = a + d$ which implies $d = 0$.
In the last step I have assumed that the "usual properties" of $+$ have already been proved.
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $mathbb{N}$ denote the set of natural numbers and $a, b in mathbb{N}$.
1) Let $a < b$. Then $b = a + d$ for some $d in mathbb{N}$ and $a ne b$. Hence $d = 0$ is impossible because $a + 0 = a$, that is, $b = a + d$ for some positive $d in mathbb{N}$.
2) Let $b = a + d$ for some positive $d in mathbb{N}$. Then certainly $a le b$. But $a = b$ is impossible because in that case $a = a + d$ which implies $d = 0$.
In the last step I have assumed that the "usual properties" of $+$ have already been proved.
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
$endgroup$
add a comment |
$begingroup$
Let $mathbb{N}$ denote the set of natural numbers and $a, b in mathbb{N}$.
1) Let $a < b$. Then $b = a + d$ for some $d in mathbb{N}$ and $a ne b$. Hence $d = 0$ is impossible because $a + 0 = a$, that is, $b = a + d$ for some positive $d in mathbb{N}$.
2) Let $b = a + d$ for some positive $d in mathbb{N}$. Then certainly $a le b$. But $a = b$ is impossible because in that case $a = a + d$ which implies $d = 0$.
In the last step I have assumed that the "usual properties" of $+$ have already been proved.
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
$endgroup$
add a comment |
$begingroup$
Let $mathbb{N}$ denote the set of natural numbers and $a, b in mathbb{N}$.
1) Let $a < b$. Then $b = a + d$ for some $d in mathbb{N}$ and $a ne b$. Hence $d = 0$ is impossible because $a + 0 = a$, that is, $b = a + d$ for some positive $d in mathbb{N}$.
2) Let $b = a + d$ for some positive $d in mathbb{N}$. Then certainly $a le b$. But $a = b$ is impossible because in that case $a = a + d$ which implies $d = 0$.
In the last step I have assumed that the "usual properties" of $+$ have already been proved.
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
$endgroup$
Let $mathbb{N}$ denote the set of natural numbers and $a, b in mathbb{N}$.
1) Let $a < b$. Then $b = a + d$ for some $d in mathbb{N}$ and $a ne b$. Hence $d = 0$ is impossible because $a + 0 = a$, that is, $b = a + d$ for some positive $d in mathbb{N}$.
2) Let $b = a + d$ for some positive $d in mathbb{N}$. Then certainly $a le b$. But $a = b$ is impossible because in that case $a = a + d$ which implies $d = 0$.
In the last step I have assumed that the "usual properties" of $+$ have already been proved.
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
answered Dec 13 '18 at 18:05
Paul FrostPaul Frost
10.5k3933
10.5k3933
add a comment |
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$begingroup$
For which set of number do you want to prove the statement? The natural numbers?
$endgroup$
– Thomas
Dec 13 '18 at 17:28
$begingroup$
Yes, the natural numbers
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:32
$begingroup$
And the natrural numbers are $0,1,2,3,dots$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:50
$begingroup$
Yes they are defined like that
$endgroup$
– Fruchtsaft
Dec 13 '18 at 17:53