Prove through induction that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|{{i|a in A_i}}| $...
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I am trying to prove the following claim through induction:
Assume $A_1,ldots,A_n$ series of Sets.
Prove that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|{{i|a in A_i}}| $ is odd
My questions:
- The base case starts here from one or from zero?
- Should I prove this claim in two ways? right to left and left to right? i.e do two inductions?
elementary-set-theory induction
asked Dec 29 '18 at 17:01
John DJohn D
977
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add a comment |
$begingroup$
I am trying to prove the following claim through induction:
Assume $A_1,ldots,A_n$ series of Sets.
Prove that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|{{i|a in A_i}}| $ is odd
My questions:
- The base case starts here from one or from zero?
- Should I prove this claim in two ways? right to left and left to right? i.e do two inductions?
elementary-set-theory induction
asked Dec 29 '18 at 17:01
John DJohn D
977
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following claim through induction:
Assume $A_1,ldots,A_n$ series of Sets.
Prove that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|{{i|a in A_i}}| $ is odd
My questions:
- The base case starts here from one or from zero?
- Should I prove this claim in two ways? right to left and left to right? i.e do two inductions?
elementary-set-theory induction
asked Dec 29 '18 at 17:01
John DJohn D
977
$endgroup$
I am trying to prove the following claim through induction:
Assume $A_1,ldots,A_n$ series of Sets.
Prove that $a in A_1 triangle A_2 triangle ldots triangle A_n $ $iff$ $|{{i|a in A_i}}| $ is odd
My questions:
- The base case starts here from one or from zero?
- Should I prove this claim in two ways? right to left and left to right? i.e do two inductions?
elementary-set-theory induction
elementary-set-theory induction
asked Dec 29 '18 at 17:01
John DJohn D
977
asked Dec 29 '18 at 17:01
John DJohn D
977
asked Dec 29 '18 at 17:01
John DJohn D
977
asked Dec 29 '18 at 17:01
John DJohn D
977
asked Dec 29 '18 at 17:01
John DJohn D
977
977
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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You can start from $0$, because one can define the multiple symmetric difference by
$$
mathop{largetriangle}_{i=0}^0 A_i=emptyset, qquad
mathop{largetriangle}_{i=0}^{n+1} A_i=
biggl(mathop{largetriangle}_{i=0}^{n} A_ibiggr)mathbin{triangle}A_{n+1}
$$
Here, associativity of symmetric difference is important.
Now the base step of the induction is clear: for $n=0$, we have $|{imid ain A_i}|=0$ and $anotinemptyset$.
Now suppose that the statement holds for $n$ sets and set, for simplicity,
$$
B=mathop{largetriangle}_{i=0}^{n} A_i
$$
We have that $ain Bmathbin{triangle}A$ if and only if $ain B$ or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$. This is equivalent, by the induction hypothesis, to
$|{imid ain A_i}|$ is odd or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$.
Check the cases and you're done.
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
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Base case is for $n=1$:
$$a in A_1 iff |{i in {1} : a in A_i}| = 1 iff |{iin {1} : a in A_i}| text{ is odd}$$
since $|{i in {1} : a in A_i}| in {0,1}$.
Assume that $a in A_1 Delta, cdots Delta,A_n iff |{i in {1, ldots, n}: a in A_i}|$ is odd.
For $n+1$ we have
begin{align}
a in A_1 Delta, cdots Delta,A_n Delta,A_{n+1} &iff veebegin{cases} big(a in A_1 Delta, cdots Delta,A_nbig) wedge (a notin A_{n+1}), \
big(a notin A_1 Delta, cdots Delta,A_nbig) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} big(|{i in {1, ldots, n}: a in A_i}| text{ is odd}big) wedge (a notin A_{n+1}), \
big(|{i in {1, ldots, n}: a in A_i}| text{ is even}big) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} |{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}, \
|{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}end{cases}\
&iff |{i in {1, ldots, n,n+1}: a in A_i}|text{ is odd}
end{align}
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
You can start from $0$, because one can define the multiple symmetric difference by
$$
mathop{largetriangle}_{i=0}^0 A_i=emptyset, qquad
mathop{largetriangle}_{i=0}^{n+1} A_i=
biggl(mathop{largetriangle}_{i=0}^{n} A_ibiggr)mathbin{triangle}A_{n+1}
$$
Here, associativity of symmetric difference is important.
Now the base step of the induction is clear: for $n=0$, we have $|{imid ain A_i}|=0$ and $anotinemptyset$.
Now suppose that the statement holds for $n$ sets and set, for simplicity,
$$
B=mathop{largetriangle}_{i=0}^{n} A_i
$$
We have that $ain Bmathbin{triangle}A$ if and only if $ain B$ or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$. This is equivalent, by the induction hypothesis, to
$|{imid ain A_i}|$ is odd or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$.
Check the cases and you're done.
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
You can start from $0$, because one can define the multiple symmetric difference by
$$
mathop{largetriangle}_{i=0}^0 A_i=emptyset, qquad
mathop{largetriangle}_{i=0}^{n+1} A_i=
biggl(mathop{largetriangle}_{i=0}^{n} A_ibiggr)mathbin{triangle}A_{n+1}
$$
Here, associativity of symmetric difference is important.
Now the base step of the induction is clear: for $n=0$, we have $|{imid ain A_i}|=0$ and $anotinemptyset$.
Now suppose that the statement holds for $n$ sets and set, for simplicity,
$$
B=mathop{largetriangle}_{i=0}^{n} A_i
$$
We have that $ain Bmathbin{triangle}A$ if and only if $ain B$ or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$. This is equivalent, by the induction hypothesis, to
$|{imid ain A_i}|$ is odd or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$.
Check the cases and you're done.
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
You can start from $0$, because one can define the multiple symmetric difference by
$$
mathop{largetriangle}_{i=0}^0 A_i=emptyset, qquad
mathop{largetriangle}_{i=0}^{n+1} A_i=
biggl(mathop{largetriangle}_{i=0}^{n} A_ibiggr)mathbin{triangle}A_{n+1}
$$
Here, associativity of symmetric difference is important.
Now the base step of the induction is clear: for $n=0$, we have $|{imid ain A_i}|=0$ and $anotinemptyset$.
Now suppose that the statement holds for $n$ sets and set, for simplicity,
$$
B=mathop{largetriangle}_{i=0}^{n} A_i
$$
We have that $ain Bmathbin{triangle}A$ if and only if $ain B$ or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$. This is equivalent, by the induction hypothesis, to
$|{imid ain A_i}|$ is odd or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$.
Check the cases and you're done.
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
$endgroup$
You can start from $0$, because one can define the multiple symmetric difference by
$$
mathop{largetriangle}_{i=0}^0 A_i=emptyset, qquad
mathop{largetriangle}_{i=0}^{n+1} A_i=
biggl(mathop{largetriangle}_{i=0}^{n} A_ibiggr)mathbin{triangle}A_{n+1}
$$
Here, associativity of symmetric difference is important.
Now the base step of the induction is clear: for $n=0$, we have $|{imid ain A_i}|=0$ and $anotinemptyset$.
Now suppose that the statement holds for $n$ sets and set, for simplicity,
$$
B=mathop{largetriangle}_{i=0}^{n} A_i
$$
We have that $ain Bmathbin{triangle}A$ if and only if $ain B$ or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$. This is equivalent, by the induction hypothesis, to
$|{imid ain A_i}|$ is odd or $ain A_{n+1}$, but $anotin Bcap A_{n+1}$.
Check the cases and you're done.
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
answered Dec 29 '18 at 17:21
egregegreg
186k1486209
186k1486209
add a comment |
add a comment |
$begingroup$
Base case is for $n=1$:
$$a in A_1 iff |{i in {1} : a in A_i}| = 1 iff |{iin {1} : a in A_i}| text{ is odd}$$
since $|{i in {1} : a in A_i}| in {0,1}$.
Assume that $a in A_1 Delta, cdots Delta,A_n iff |{i in {1, ldots, n}: a in A_i}|$ is odd.
For $n+1$ we have
begin{align}
a in A_1 Delta, cdots Delta,A_n Delta,A_{n+1} &iff veebegin{cases} big(a in A_1 Delta, cdots Delta,A_nbig) wedge (a notin A_{n+1}), \
big(a notin A_1 Delta, cdots Delta,A_nbig) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} big(|{i in {1, ldots, n}: a in A_i}| text{ is odd}big) wedge (a notin A_{n+1}), \
big(|{i in {1, ldots, n}: a in A_i}| text{ is even}big) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} |{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}, \
|{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}end{cases}\
&iff |{i in {1, ldots, n,n+1}: a in A_i}|text{ is odd}
end{align}
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
Base case is for $n=1$:
$$a in A_1 iff |{i in {1} : a in A_i}| = 1 iff |{iin {1} : a in A_i}| text{ is odd}$$
since $|{i in {1} : a in A_i}| in {0,1}$.
Assume that $a in A_1 Delta, cdots Delta,A_n iff |{i in {1, ldots, n}: a in A_i}|$ is odd.
For $n+1$ we have
begin{align}
a in A_1 Delta, cdots Delta,A_n Delta,A_{n+1} &iff veebegin{cases} big(a in A_1 Delta, cdots Delta,A_nbig) wedge (a notin A_{n+1}), \
big(a notin A_1 Delta, cdots Delta,A_nbig) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} big(|{i in {1, ldots, n}: a in A_i}| text{ is odd}big) wedge (a notin A_{n+1}), \
big(|{i in {1, ldots, n}: a in A_i}| text{ is even}big) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} |{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}, \
|{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}end{cases}\
&iff |{i in {1, ldots, n,n+1}: a in A_i}|text{ is odd}
end{align}
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
$endgroup$
add a comment |
$begingroup$
Base case is for $n=1$:
$$a in A_1 iff |{i in {1} : a in A_i}| = 1 iff |{iin {1} : a in A_i}| text{ is odd}$$
since $|{i in {1} : a in A_i}| in {0,1}$.
Assume that $a in A_1 Delta, cdots Delta,A_n iff |{i in {1, ldots, n}: a in A_i}|$ is odd.
For $n+1$ we have
begin{align}
a in A_1 Delta, cdots Delta,A_n Delta,A_{n+1} &iff veebegin{cases} big(a in A_1 Delta, cdots Delta,A_nbig) wedge (a notin A_{n+1}), \
big(a notin A_1 Delta, cdots Delta,A_nbig) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} big(|{i in {1, ldots, n}: a in A_i}| text{ is odd}big) wedge (a notin A_{n+1}), \
big(|{i in {1, ldots, n}: a in A_i}| text{ is even}big) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} |{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}, \
|{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}end{cases}\
&iff |{i in {1, ldots, n,n+1}: a in A_i}|text{ is odd}
end{align}
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
$endgroup$
Base case is for $n=1$:
$$a in A_1 iff |{i in {1} : a in A_i}| = 1 iff |{iin {1} : a in A_i}| text{ is odd}$$
since $|{i in {1} : a in A_i}| in {0,1}$.
Assume that $a in A_1 Delta, cdots Delta,A_n iff |{i in {1, ldots, n}: a in A_i}|$ is odd.
For $n+1$ we have
begin{align}
a in A_1 Delta, cdots Delta,A_n Delta,A_{n+1} &iff veebegin{cases} big(a in A_1 Delta, cdots Delta,A_nbig) wedge (a notin A_{n+1}), \
big(a notin A_1 Delta, cdots Delta,A_nbig) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} big(|{i in {1, ldots, n}: a in A_i}| text{ is odd}big) wedge (a notin A_{n+1}), \
big(|{i in {1, ldots, n}: a in A_i}| text{ is even}big) wedge (a in A_{n+1})end{cases}\
&iff veebegin{cases} |{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}, \
|{i in {1, ldots, n,n+1}: a in A_i}| text{ is odd}end{cases}\
&iff |{i in {1, ldots, n,n+1}: a in A_i}|text{ is odd}
end{align}
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
answered Dec 29 '18 at 17:19
mechanodroidmechanodroid
28.9k62648
28.9k62648
add a comment |
add a comment |
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