Discrete Math - Proving Distributive Laws for Sets by induction
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I'm working on doing a proof by induction on this question:
Use induction to prove that if $X_1, . . . , X_n$ and $X$ are sets, then
$X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$.
I've shown the basis case:
$X∩X_1 = X∩X_1$
But I'm having difficulty proving the $n + 1$ case. This is my work so far:
Assume $X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$
Show $X∩(X_1∪X_2∪· · ·∪X_n∪X_{n+1}) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)∪(X∩X_{n+1})$
I'm not sure what I'm allowed to do from here. Where can I use the induction hypothesis?
discrete-mathematics elementary-set-theory induction proof-explanation
edited Feb 19 at 11:41
Heptapod
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
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add a comment |
$begingroup$
I'm working on doing a proof by induction on this question:
Use induction to prove that if $X_1, . . . , X_n$ and $X$ are sets, then
$X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$.
I've shown the basis case:
$X∩X_1 = X∩X_1$
But I'm having difficulty proving the $n + 1$ case. This is my work so far:
Assume $X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$
Show $X∩(X_1∪X_2∪· · ·∪X_n∪X_{n+1}) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)∪(X∩X_{n+1})$
I'm not sure what I'm allowed to do from here. Where can I use the induction hypothesis?
discrete-mathematics elementary-set-theory induction proof-explanation
edited Feb 19 at 11:41
Heptapod
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
$endgroup$
$begingroup$
Please use MathJax to format your question correctly
$endgroup$
– Ankit Kumar
Jan 11 at 4:57
add a comment |
$begingroup$
I'm working on doing a proof by induction on this question:
Use induction to prove that if $X_1, . . . , X_n$ and $X$ are sets, then
$X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$.
I've shown the basis case:
$X∩X_1 = X∩X_1$
But I'm having difficulty proving the $n + 1$ case. This is my work so far:
Assume $X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$
Show $X∩(X_1∪X_2∪· · ·∪X_n∪X_{n+1}) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)∪(X∩X_{n+1})$
I'm not sure what I'm allowed to do from here. Where can I use the induction hypothesis?
discrete-mathematics elementary-set-theory induction proof-explanation
edited Feb 19 at 11:41
Heptapod
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
$endgroup$
I'm working on doing a proof by induction on this question:
Use induction to prove that if $X_1, . . . , X_n$ and $X$ are sets, then
$X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$.
I've shown the basis case:
$X∩X_1 = X∩X_1$
But I'm having difficulty proving the $n + 1$ case. This is my work so far:
Assume $X∩(X_1∪X_2∪· · ·∪X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)$
Show $X∩(X_1∪X_2∪· · ·∪X_n∪X_{n+1}) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n) = (X∩X_1)∪(X∩X_2)∪· · ·∪(X∩X_n)∪(X∩X_{n+1})$
I'm not sure what I'm allowed to do from here. Where can I use the induction hypothesis?
discrete-mathematics elementary-set-theory induction proof-explanation
discrete-mathematics elementary-set-theory induction proof-explanation
edited Feb 19 at 11:41
Heptapod
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
edited Feb 19 at 11:41
Heptapod
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
edited Feb 19 at 11:41
Heptapod
716417
edited Feb 19 at 11:41
Heptapod
716417
edited Feb 19 at 11:41
Heptapod
716417
716417
asked Jan 11 at 4:49
ShellG44ShellG44
62
asked Jan 11 at 4:49
ShellG44ShellG44
62
asked Jan 11 at 4:49
ShellG44ShellG44
62
62
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Please use MathJax to format your question correctly
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– Ankit Kumar
Jan 11 at 4:57
add a comment |
$begingroup$
Please use MathJax to format your question correctly
$endgroup$
– Ankit Kumar
Jan 11 at 4:57
$begingroup$
Please use MathJax to format your question correctly
$endgroup$
– Ankit Kumar
Jan 11 at 4:57
$begingroup$
Please use MathJax to format your question correctly
$endgroup$
– Ankit Kumar
Jan 11 at 4:57
add a comment |
1 Answer
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Tip show: $Xcap(Ycup X_{n+1})=(Xcup Y)cap(Xcup X_{n+1})$ where $Y=X_1cupcdotscup X_n$
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
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1 Answer
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1 Answer
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$begingroup$
Tip show: $Xcap(Ycup X_{n+1})=(Xcup Y)cap(Xcup X_{n+1})$ where $Y=X_1cupcdotscup X_n$
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
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add a comment |
$begingroup$
Tip show: $Xcap(Ycup X_{n+1})=(Xcup Y)cap(Xcup X_{n+1})$ where $Y=X_1cupcdotscup X_n$
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
$endgroup$
add a comment |
$begingroup$
Tip show: $Xcap(Ycup X_{n+1})=(Xcup Y)cap(Xcup X_{n+1})$ where $Y=X_1cupcdotscup X_n$
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
$endgroup$
Tip show: $Xcap(Ycup X_{n+1})=(Xcup Y)cap(Xcup X_{n+1})$ where $Y=X_1cupcdotscup X_n$
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
answered Jan 11 at 4:54
Graham KempGraham Kemp
87.9k43578
87.9k43578
add a comment |
add a comment |
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$begingroup$
Please use MathJax to format your question correctly
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– Ankit Kumar
Jan 11 at 4:57