Lemma 2.2.3. For any natural numbers $n$ and $m$, $n + ( m{+}+)$ = $(n+m){+}+$: Analysis 1, Terence Tao
$begingroup$
I'm self-learning real analysis using Terence Tao Analysis books. The book is very lucid. But in some cases Terence quickly go through the proof. And it becomes difficult for lesser mortals like us to understand. Lemma 2.2.3 is one of them. So, I'm requesting you guyz if you can help me understand this proof. The proof is the following:
The portion, I'm not getting is:
But by definition of addition, $0 + (m{+}+) = m{+}+$ and $0 + m = m$, so both sides are equal to $m{+}+$ and are thus equal to each other.
Why $m{+}+$ & $m$ are equal. Similay, he is using similar logic in following section:
Similarly, we have $(n{+}+ )+m = (n+m){+}+$ by the definition of addition, and so the right-hand side is also equal to $((n+m){+}+){+}+.$
These might be very simple logic. But I'm having hard time getting grasp of it. If someone can explain it, I'll be extremely grateful.
Thanks!
real-analysis induction
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
$endgroup$
add a comment |
$begingroup$
I'm self-learning real analysis using Terence Tao Analysis books. The book is very lucid. But in some cases Terence quickly go through the proof. And it becomes difficult for lesser mortals like us to understand. Lemma 2.2.3 is one of them. So, I'm requesting you guyz if you can help me understand this proof. The proof is the following:
The portion, I'm not getting is:
But by definition of addition, $0 + (m{+}+) = m{+}+$ and $0 + m = m$, so both sides are equal to $m{+}+$ and are thus equal to each other.
Why $m{+}+$ & $m$ are equal. Similay, he is using similar logic in following section:
Similarly, we have $(n{+}+ )+m = (n+m){+}+$ by the definition of addition, and so the right-hand side is also equal to $((n+m){+}+){+}+.$
These might be very simple logic. But I'm having hard time getting grasp of it. If someone can explain it, I'll be extremely grateful.
Thanks!
real-analysis induction
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
$endgroup$
add a comment |
$begingroup$
I'm self-learning real analysis using Terence Tao Analysis books. The book is very lucid. But in some cases Terence quickly go through the proof. And it becomes difficult for lesser mortals like us to understand. Lemma 2.2.3 is one of them. So, I'm requesting you guyz if you can help me understand this proof. The proof is the following:
The portion, I'm not getting is:
But by definition of addition, $0 + (m{+}+) = m{+}+$ and $0 + m = m$, so both sides are equal to $m{+}+$ and are thus equal to each other.
Why $m{+}+$ & $m$ are equal. Similay, he is using similar logic in following section:
Similarly, we have $(n{+}+ )+m = (n+m){+}+$ by the definition of addition, and so the right-hand side is also equal to $((n+m){+}+){+}+.$
These might be very simple logic. But I'm having hard time getting grasp of it. If someone can explain it, I'll be extremely grateful.
Thanks!
real-analysis induction
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
$endgroup$
I'm self-learning real analysis using Terence Tao Analysis books. The book is very lucid. But in some cases Terence quickly go through the proof. And it becomes difficult for lesser mortals like us to understand. Lemma 2.2.3 is one of them. So, I'm requesting you guyz if you can help me understand this proof. The proof is the following:
The portion, I'm not getting is:
But by definition of addition, $0 + (m{+}+) = m{+}+$ and $0 + m = m$, so both sides are equal to $m{+}+$ and are thus equal to each other.
Why $m{+}+$ & $m$ are equal. Similay, he is using similar logic in following section:
Similarly, we have $(n{+}+ )+m = (n+m){+}+$ by the definition of addition, and so the right-hand side is also equal to $((n+m){+}+){+}+.$
These might be very simple logic. But I'm having hard time getting grasp of it. If someone can explain it, I'll be extremely grateful.
Thanks!
real-analysis induction
real-analysis induction
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
edited Dec 17 '18 at 15:34
Christoph
12.1k1642
12.1k1642
asked Dec 17 '18 at 15:15
BetaBeta
1256
asked Dec 17 '18 at 15:15
BetaBeta
1256
asked Dec 17 '18 at 15:15
BetaBeta
1256
1256
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Tao never asserts that $m$ and $m++$ are equal. What he asserts is that $0+(m{+}+)=m{+}+$ and that $(0+m){+}+=m{+}+$. So, in particular, since both $0+(m{+}+)$ and $(0+m){+}+$ are equal to $m{+}+$, $0+(m{+}+)=(0+m){+}+$.
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please usem{+}+instead ofm++in MathJax.
$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
|
show 1 more comment
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Tao never asserts that $m$ and $m++$ are equal. What he asserts is that $0+(m{+}+)=m{+}+$ and that $(0+m){+}+=m{+}+$. So, in particular, since both $0+(m{+}+)$ and $(0+m){+}+$ are equal to $m{+}+$, $0+(m{+}+)=(0+m){+}+$.
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please usem{+}+instead ofm++in MathJax.
$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
|
show 1 more comment
$begingroup$
Tao never asserts that $m$ and $m++$ are equal. What he asserts is that $0+(m{+}+)=m{+}+$ and that $(0+m){+}+=m{+}+$. So, in particular, since both $0+(m{+}+)$ and $(0+m){+}+$ are equal to $m{+}+$, $0+(m{+}+)=(0+m){+}+$.
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please usem{+}+instead ofm++in MathJax.
$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
|
show 1 more comment
$begingroup$
Tao never asserts that $m$ and $m++$ are equal. What he asserts is that $0+(m{+}+)=m{+}+$ and that $(0+m){+}+=m{+}+$. So, in particular, since both $0+(m{+}+)$ and $(0+m){+}+$ are equal to $m{+}+$, $0+(m{+}+)=(0+m){+}+$.
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
Tao never asserts that $m$ and $m++$ are equal. What he asserts is that $0+(m{+}+)=m{+}+$ and that $(0+m){+}+=m{+}+$. So, in particular, since both $0+(m{+}+)$ and $(0+m){+}+$ are equal to $m{+}+$, $0+(m{+}+)=(0+m){+}+$.
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
edited Dec 17 '18 at 15:35
Christoph
12.1k1642
12.1k1642
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 17 '18 at 15:25
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please usem{+}+instead ofm++in MathJax.
$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
|
show 1 more comment
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please usem{+}+instead ofm++in MathJax.
$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
Thanks for your answer! This clears the doubt. But in the line "But by definition of addition, 0 + (m++) = m++ and 0 + m = m", why he added "0+m = m". We know by definition 0 + (m++) = m++. What's the reason for adding additional equation? Your answer completely explained what I asked. But just wanted to be thorough in my understanding. Thanks again!
$endgroup$
– Beta
Dec 17 '18 at 15:35
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
He needed to know that $0+m=m$ in order to deduce that $(0+m)++=m++$.
$endgroup$
– José Carlos Santos
Dec 17 '18 at 15:37
$begingroup$
For reasons of spacing, please use
m{+}+ instead of m++ in MathJax.$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
For reasons of spacing, please use
m{+}+ instead of m++ in MathJax.$endgroup$
– Christoph
Dec 17 '18 at 15:38
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@JoséCarlosSantos: Perfect!! Thanks for your answer! Each step matter in maths :)
$endgroup$
– Beta
Dec 17 '18 at 15:39
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
$begingroup$
@Christoph: Sure...will do this from the next time.
$endgroup$
– Beta
Dec 17 '18 at 15:40
|
show 1 more comment
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