Given two formulas $A$ and $B$, if $A$ follows from $B$ and $B$ follows from $A$ then is it true that $A$ and...
$begingroup$
This is true if $A$ and $B$ are statements, but formulas are statements too, so I expect the answer to be yes. But let's consider this simple example:
Formula A: $$sum_{k=0}^{n-1}x^k=dfrac{1-x^n}{1-x}$$
Formula B: $$sum_{k=0}^infty x^k=dfrac{1}{1-x},,,,, |x|<1$$
Proof that $Ato B$: Just take the limit $nto infty $ in $A$ assuming $|x|<1$.
Proof that $Bto A$: $$sum_{k=0}^infty x^k=sum_{k=0}^{n-1}sum_{m=0}^infty x^{k+mn} = sum_{k=0}^{n-1}x^kcdot sum_{m=0}^infty x^{mn}$$
From this using $B$ we get $$dfrac{1}{1-x}=sum_{k=0}^{n-1}x^kcdot dfrac{1}{1-x^n},,,,, |x|<1 $$ So we get that $A$ is true if $|x|<1$. But the condition $|x|<1$ can be dropped because this is a polynomial relation, thus $A$ follows.
But clearly $A$ and $B$ are not equivalent. Am I missing something here?
real-analysis logic recreational-mathematics meta-math
asked Jan 6 at 13:22
TyrellTyrell
587321
$endgroup$
add a comment |
$begingroup$
This is true if $A$ and $B$ are statements, but formulas are statements too, so I expect the answer to be yes. But let's consider this simple example:
Formula A: $$sum_{k=0}^{n-1}x^k=dfrac{1-x^n}{1-x}$$
Formula B: $$sum_{k=0}^infty x^k=dfrac{1}{1-x},,,,, |x|<1$$
Proof that $Ato B$: Just take the limit $nto infty $ in $A$ assuming $|x|<1$.
Proof that $Bto A$: $$sum_{k=0}^infty x^k=sum_{k=0}^{n-1}sum_{m=0}^infty x^{k+mn} = sum_{k=0}^{n-1}x^kcdot sum_{m=0}^infty x^{mn}$$
From this using $B$ we get $$dfrac{1}{1-x}=sum_{k=0}^{n-1}x^kcdot dfrac{1}{1-x^n},,,,, |x|<1 $$ So we get that $A$ is true if $|x|<1$. But the condition $|x|<1$ can be dropped because this is a polynomial relation, thus $A$ follows.
But clearly $A$ and $B$ are not equivalent. Am I missing something here?
real-analysis logic recreational-mathematics meta-math
asked Jan 6 at 13:22
TyrellTyrell
587321
$endgroup$
2
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02
add a comment |
$begingroup$
This is true if $A$ and $B$ are statements, but formulas are statements too, so I expect the answer to be yes. But let's consider this simple example:
Formula A: $$sum_{k=0}^{n-1}x^k=dfrac{1-x^n}{1-x}$$
Formula B: $$sum_{k=0}^infty x^k=dfrac{1}{1-x},,,,, |x|<1$$
Proof that $Ato B$: Just take the limit $nto infty $ in $A$ assuming $|x|<1$.
Proof that $Bto A$: $$sum_{k=0}^infty x^k=sum_{k=0}^{n-1}sum_{m=0}^infty x^{k+mn} = sum_{k=0}^{n-1}x^kcdot sum_{m=0}^infty x^{mn}$$
From this using $B$ we get $$dfrac{1}{1-x}=sum_{k=0}^{n-1}x^kcdot dfrac{1}{1-x^n},,,,, |x|<1 $$ So we get that $A$ is true if $|x|<1$. But the condition $|x|<1$ can be dropped because this is a polynomial relation, thus $A$ follows.
But clearly $A$ and $B$ are not equivalent. Am I missing something here?
real-analysis logic recreational-mathematics meta-math
asked Jan 6 at 13:22
TyrellTyrell
587321
$endgroup$
This is true if $A$ and $B$ are statements, but formulas are statements too, so I expect the answer to be yes. But let's consider this simple example:
Formula A: $$sum_{k=0}^{n-1}x^k=dfrac{1-x^n}{1-x}$$
Formula B: $$sum_{k=0}^infty x^k=dfrac{1}{1-x},,,,, |x|<1$$
Proof that $Ato B$: Just take the limit $nto infty $ in $A$ assuming $|x|<1$.
Proof that $Bto A$: $$sum_{k=0}^infty x^k=sum_{k=0}^{n-1}sum_{m=0}^infty x^{k+mn} = sum_{k=0}^{n-1}x^kcdot sum_{m=0}^infty x^{mn}$$
From this using $B$ we get $$dfrac{1}{1-x}=sum_{k=0}^{n-1}x^kcdot dfrac{1}{1-x^n},,,,, |x|<1 $$ So we get that $A$ is true if $|x|<1$. But the condition $|x|<1$ can be dropped because this is a polynomial relation, thus $A$ follows.
But clearly $A$ and $B$ are not equivalent. Am I missing something here?
real-analysis logic recreational-mathematics meta-math
real-analysis logic recreational-mathematics meta-math
asked Jan 6 at 13:22
TyrellTyrell
587321
asked Jan 6 at 13:22
TyrellTyrell
587321
asked Jan 6 at 13:22
TyrellTyrell
587321
asked Jan 6 at 13:22
TyrellTyrell
587321
asked Jan 6 at 13:22
TyrellTyrell
587321
587321
2
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02
add a comment |
2
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02
2
2
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02
add a comment |
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2
$begingroup$
"But clearly $A$ and $B$ are not equivalent". Why not? They're both true, and anything that is true is equivalent to anything else that is true.
$endgroup$
– Henning Makholm
Jan 6 at 13:45
$begingroup$
A does not imply B. Add for all n in N to A and then it will imply B.
$endgroup$
– William Elliot
Jan 6 at 13:48
$begingroup$
@WilliamElliot: $A$ does imply $B$ in any of the obvious contexts for interpreting the question (e.g., in $Bbb{R}$ or $Bbb{C}$ with the usual notion of limit). What context do you have in mind where $A$ will not imply $B$?
$endgroup$
– Rob Arthan
Jan 8 at 21:02