If $(a_n)to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$Proving $b_n$$_(n=1)^infty$ converges given $a_n$$_n=1^infty$ and $a_n + b_n$$_n=1^infty$ convergeProve that sequences $fraca_nb_n = 0$Show $ a_n-b_n rightarrow 0$ then $b_n rightarrow L$$c_n$ is a shuffling of $a_n$ and $b_n$. Prove that $c_n$ converges iff $a_n$ and $b_n$ converge to the same number.If $a_n$ and $ a_n+ b_n $ are convergent sequences, then $ b_n $ convergesIf $a_n$ and $a_nb_n$ are convergent sequences, then $b_n$ converges.If $a_n$ and $b_n$ converge, then $a_nb_n$ converges.Prove that if $M neq 0$, then $fraca_nb_n Rightarrow fracLM$If $a_n$ and $b_n$ are converging sequences, and for all $ninmathbb Z+$, $a_n leq b_n$, then $lim_ntoinfty a_n leq lim_ntoinfty b_n$.Given that $a_n$ is a sequence of $mathbbR$ s.t. $a_n leq b_n$ and $a_n rightarrow a$ and $b_n rightarrow b$ then $a leq b$
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If $(a_n)to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$
Proving $b_n$$_(n=1)^infty$ converges given $a_n$$_n=1^infty$ and $a_n + b_n$$_n=1^infty$ convergeProve that sequences $fraca_nb_n = 0$Show $ a_n-b_n rightarrow 0$ then $b_n rightarrow L$$c_n$ is a shuffling of $a_n$ and $b_n$. Prove that $c_n$ converges iff $a_n$ and $b_n$ converge to the same number.If $a_n$ and $ a_n+ b_n $ are convergent sequences, then $ b_n $ convergesIf $a_n$ and $a_nb_n$ are convergent sequences, then $b_n$ converges.If $a_n$ and $b_n$ converge, then $a_nb_n$ converges.Prove that if $M neq 0$, then $fraca_nb_n Rightarrow fracLM$If $a_n$ and $b_n$ are converging sequences, and for all $ninmathbb Z+$, $a_n leq b_n$, then $lim_ntoinfty a_n leq lim_ntoinfty b_n$.Given that $a_n$ is a sequence of $mathbbR$ s.t. $a_n leq b_n$ and $a_n rightarrow a$ and $b_n rightarrow b$ then $a leq b$
$begingroup$
Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$
I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that
$$|b_n - B| < epsilon$$
I know that since $<a_n>to A$ then there exists an $N_1in mathbbN$ such that for all $n > N_1$
$$|a_n - A| < epsilon$$
Similarly, since $<a_n b_n>to AB$ then there exists an $N_2inmathbbN$ such that for all $n > N_2$
$$|a_n b_n - AB| < epsilon$$
Any help would be appreciated.
Background information:
Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$
Lemma 8.1 - If $a_nneq 0$ for all $ninmathbbN$ and if $(a_n)$ converges to $Aneq 0$ then the sequence $(1/a_n)$ converges to $1/A$
Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.
I don't think this is complete can someone help me add details?
real-analysis sequences-and-series proof-writing
edited Mar 29 at 4:22
dmtri
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
add a comment |
$begingroup$
Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$
I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that
$$|b_n - B| < epsilon$$
I know that since $<a_n>to A$ then there exists an $N_1in mathbbN$ such that for all $n > N_1$
$$|a_n - A| < epsilon$$
Similarly, since $<a_n b_n>to AB$ then there exists an $N_2inmathbbN$ such that for all $n > N_2$
$$|a_n b_n - AB| < epsilon$$
Any help would be appreciated.
Background information:
Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$
Lemma 8.1 - If $a_nneq 0$ for all $ninmathbbN$ and if $(a_n)$ converges to $Aneq 0$ then the sequence $(1/a_n)$ converges to $1/A$
Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.
I don't think this is complete can someone help me add details?
real-analysis sequences-and-series proof-writing
edited Mar 29 at 4:22
dmtri
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
1
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
1
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
1
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11
add a comment |
$begingroup$
Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$
I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that
$$|b_n - B| < epsilon$$
I know that since $<a_n>to A$ then there exists an $N_1in mathbbN$ such that for all $n > N_1$
$$|a_n - A| < epsilon$$
Similarly, since $<a_n b_n>to AB$ then there exists an $N_2inmathbbN$ such that for all $n > N_2$
$$|a_n b_n - AB| < epsilon$$
Any help would be appreciated.
Background information:
Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$
Lemma 8.1 - If $a_nneq 0$ for all $ninmathbbN$ and if $(a_n)$ converges to $Aneq 0$ then the sequence $(1/a_n)$ converges to $1/A$
Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.
I don't think this is complete can someone help me add details?
real-analysis sequences-and-series proof-writing
edited Mar 29 at 4:22
dmtri
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
$endgroup$
Let $(a_n)$ and $(b_n)$ be sequences. If $(a_n) to Aneq 0$ and $(a_n b_n)to AB$ then $(b_n)to B$
I know that we need to show this for both $A > 0$ and $A < 0$. But I am having problems using the assumptions to deduce that
$$|b_n - B| < epsilon$$
I know that since $<a_n>to A$ then there exists an $N_1in mathbbN$ such that for all $n > N_1$
$$|a_n - A| < epsilon$$
Similarly, since $<a_n b_n>to AB$ then there exists an $N_2inmathbbN$ such that for all $n > N_2$
$$|a_n b_n - AB| < epsilon$$
Any help would be appreciated.
Background information:
Theorem 8.7 - If the sequence $(a_n)$ converges to $A$ and the sequence $(b_n)$ converges to $B$ then the sequence $(a_n b _n)$ converges to $AB$
Lemma 8.1 - If $a_nneq 0$ for all $ninmathbbN$ and if $(a_n)$ converges to $Aneq 0$ then the sequence $(1/a_n)$ converges to $1/A$
Attempted proof - Using these two theorems above we can write $(a_n b_n/a_n) = (b_n)$ which converges to $B$.
I don't think this is complete can someone help me add details?
real-analysis sequences-and-series proof-writing
real-analysis sequences-and-series proof-writing
edited Mar 29 at 4:22
dmtri
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
edited Mar 29 at 4:22
dmtri
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
edited Mar 29 at 4:22
dmtri
1,7712521
edited Mar 29 at 4:22
dmtri
1,7712521
edited Mar 29 at 4:22
dmtri
1,7712521
1,7712521
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
asked Mar 29 at 3:50
SnorrlaxxxSnorrlaxxx
1706
1706
1
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
1
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
1
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11
add a comment |
1
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
1
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
1
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11
1
1
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
1
1
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
1
1
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
$endgroup$
add a comment |
$begingroup$
Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.
Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.
Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.
Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
$endgroup$
add a comment |
$begingroup$
Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
$endgroup$
add a comment |
$begingroup$
Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
$endgroup$
Your proof is correct but I'd recommend justifying that $a_n neq 0$ for all $n$ sufficiently large. Note that this is necessary to make sense of the expression
$$
b_n = left( fraca_nb_na_n right).
$$
To justify that $a_n neq 0$ for all $n$ sufficiently large, we will use that $A = lima_n neq 0$. Because $a_n to A$, there exists $N in mathbbN$ such that
$$
|a_n - A| < epsilon = frac2
$$
for all $n geq N$. Therefore, for all such $n$,
beginalign*
|A| - |a_n| leq leftvert a_n - A rightvert< frac2
endalign*
whence $$
0 < frac2 < |a_n|, quad forall ngeq N.
$$
This means that the $N$-tail of $(a_n)$ is never zero and consequently the sequence
$$
fracb_na_n
$$
is well defined for all $n geq N$ (and we only really care about the tail of a sequence when discussing limits). Then, by invoking the theorem and lemma you've cited, it follows that
beginalign*
limb_n = limleft(fraca_n b_na_n right)
= fraclim(a_nb_n)lima_n
= fracABA = B.
endalign*
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
edited Mar 29 at 4:42
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
answered Mar 29 at 4:37
rolandcyprolandcyp
2,234421
2,234421
add a comment |
add a comment |
$begingroup$
Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.
Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.
Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.
Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.
Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.
Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.
Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.
Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.
Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.
Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
$endgroup$
Note that if $a_n to a$ then the sequence is bounded, that is, there is some $L$ such that $|a_n| le L$.
Suppose $b_n to b$ then $|a_nb_n-ab| le |a_n b_n -a_n b| + |a_n b - ab| le L|b_n-b|+|b||a_n-a|$. Hence $a_nb_n to ab$.
Suppose $a neq 0$. Since $a_n to a$, we see that $|a_n| ge 1over 2 |a|$ for $n$ sufficiently large (and hence non zero). Then $|1 over a - 1over a_n| = le ^2 |a_n-a|$ and hence $1 over a_n to 1 over a$.
Hence if $a_nb_n to ab$ and $a_n to a neq 0$ then
$a_n b_n 1 over a_n = b_n to ab 1 over a = b$.
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
answered Mar 29 at 4:58
copper.hatcopper.hat
128k561161
128k561161
add a comment |
add a comment |
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1
$begingroup$
Can you show that if $a_n to a, b_n to b$ then $a_n b_n to ab$? Can you show that if $a neq 0$ then $1 over a_n to 1 over a$? If so, you can combine these facts to get the above result.
$endgroup$
– copper.hat
Mar 29 at 3:55
$begingroup$
@copper.hat Using those two facts can we say that since $<a_n b_n/a_n> = <b_n>$ then $<b_n>$ converges to $B$?
$endgroup$
– Snorrlaxxx
Mar 29 at 3:57
1
$begingroup$
Yes. $$
$endgroup$
– copper.hat
Mar 29 at 4:03
1
$begingroup$
Okay, I will edit with an attempted proof.
$endgroup$
– Snorrlaxxx
Mar 29 at 4:04
$begingroup$
@copper.hat I am struggling a bit with the details
$endgroup$
– Snorrlaxxx
Mar 29 at 4:11