Closed Walk
Note 1: ETH::A&D
Note Type: Horvath Reverso
GUID:
A1y[0:/g)f
Before
Front
Closed Walk
Back
Closed Walk
Zyklus
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | Closed Walk | |
| Back | Zyklus |
Note 2: ETH::A&D
Note Type: Horvath Reverso
GUID:
B+m&Yt;~bM
Before
Front
Path
Back
Path
Pfad
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | Path | |
| Back | Pfad |
Note 3: ETH::A&D
Note Type: Horvath Cloze
GUID:
C}:U@+B*;Q
Before
Front
Back
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | {{c1:: \(\sum_{i = 1}^{n} i^3\)}} \(\leq\) {{c2::\(O(n^4)\) |
{{c1:: \(\sum_{i = 1}^{n} i^3\)::Sum}} \(\leq\) {{c2::\(O(n^4)\)}} |
Note 4: ETH::A&D
Note Type: Horvath Cloze
GUID:
NU;6ob<^n3
Before
Front
Back
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | {{c1:: \(\sum_{i = 1}^{n} \sum_{k = 1}^{\textbf{i} } 1\)}} \(=\) {{c2:: \(\sum_{i = 1}^n i = \frac{n(n + 1)}{2}\) |
{{c1:: \(\sum_{i = 1}^{n} \sum_{k = 1}^{\textbf{i} } 1\)::Sum}} \(=\) {{c2:: \(\sum_{i = 1}^n i = \frac{n(n + 1)}{2}\)}} |
Note 5: ETH::A&D
Note Type: Horvath Reverso
GUID:
OW(TL-EP^P
Before
Front
Cycle
Back
Cycle
Kreis
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | Cycle | |
| Back | Kreis |
Note 6: ETH::A&D
Note Type: Horvath Reverso
GUID:
r8fq,6tx}{
Before
Front
Walk
Back
Walk
Weg
After
Front
Back
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Front | Walk | |
| Back | Weg |
Note 7: ETH::DiskMat
Note Type: Horvath Cloze
GUID:
C18gm]huq&
Before
Front
In a Group:
{{c1:: \(\widehat{a * b}\) }} = {{c2:: \(\widehat{b} * \widehat{a}\)}} This is a property from a Lemma.
Back
In a Group:
{{c1:: \(\widehat{a * b}\) }} = {{c2:: \(\widehat{b} * \widehat{a}\)}} This is a property from a Lemma.
After
Front
{{c1:: \(\widehat{a * b}\) }} = {{c2:: \(\widehat{b} * \widehat{a}\)}}
Back
{{c1:: \(\widehat{a * b}\) }} = {{c2:: \(\widehat{b} * \widehat{a}\)}}
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | In a group: <br><br>{{c1:: \(\widehat{a * b}\) }} = {{c2:: \(\widehat{b} * \widehat{a}\)}} | |
| Extra | This is a property from a Lemma. |
Note 8: ETH::DiskMat
Note Type: Horvath Classic
GUID:
k}1~03snwg
Before
Front
In a ring, \(d\) is a gcd of \(a\) and \(b\) if:
Back
In a ring, \(d\) is a gcd of \(a\) and \(b\) if:
For any ring elements \(a\) and \(b\) in \(R\) (not both \(0\)), a ring element \(d\) is called a greatest common divisor of \(a\) and \(b\) if:
- \(d\) divides both \(a\) and \(a\)0
- Every common divisor of \(a\)1 and \(a\)2 divides \(a\)3
Formally: \[d \ | \ a \ \land \ d \ | \ b \ \land \ \forall c ((c \ | \ a \ \land \ c \ | \ b) \rightarrow c \ | \ d)\]
After
Front
In a ring, \(d\) is a gcd of \(a\) and \(b\) if:
Back
In a ring, \(d\) is a gcd of \(a\) and \(b\) if:
For any ring elements \(a\) and \(b\) in \(R\) (not both \(0\)), a ring element \(d\) is called a greatest common divisor of \(a\) and \(b\) if:
- \(d\) divides both \(a\) and \(b\)
- Every common divisor of \(a\) and \(b\) divides \(d\)
Formally:\[d \ | \ a \ \land \ d \ | \ b \ \land \ \forall c ((c \ | \ a \ \land \ c \ | \ b) \rightarrow c \ | \ d)\]
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | <p>For any ring elements \(a\) and \(b\) in \(R\) (not both \(0\)), a ring element \(d\) is called a greatest common divisor of \(a\) and \(b\) if:<br>
- \(d\) divides both \(a\) and |
<p>For any ring elements \(a\) and \(b\) in \(R\) (not both \(0\)), a ring element \(d\) is called a greatest common divisor of \(a\) and \(b\) if:<br> - \(d\) divides both \(a\) and \(b\)<br> - Every common divisor of \(a\) and \(b\) divides \(d\) </p><p>Formally:\[d \ | \ a \ \land \ d \ | \ b \ \land \ \forall c ((c \ | \ a \ \land \ c \ | \ b) \rightarrow c \ | \ d)\]<br></p> |
Note 9: ETH::DiskMat
Note Type: Horvath Classic
GUID:
qAyHDnFN7L
Before
Front
Back
2. If \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by lemma)
3. The number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of coprime elements of the group.
After
Front
Back
2. If \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by Lemma)
3. The number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of elements coprime to the group order.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | 1. Verify that \(\mathbb{Z}_n^*\)is cyclic (iff n = 2, 4, \(p^e\), \(2p^e\), with \(e \ge 1\) and \(p\) is an odd prime)<br>2. If \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by |
1. Verify that \(\mathbb{Z}_n^*\)is cyclic (iff n = 2, 4, \(p^e\), \(2p^e\), with \(e \ge 1\) and \(p\) is an odd prime)<br>2. If \(\mathbb{Z}_n^*\) is cyclic then it is isomorphic to \(\mathbb{Z}_{\varphi(n)}^+\) (by Lemma) <br>3. The number of generators of \(\mathbb{Z}_{\varphi(n)}^+\) is \(\varphi(\varphi(n))\) as it is the number of elements coprime to the group order. |
Note 10: ETH::DiskMat
Note Type: Horvath Classic
GUID:
wY#5P^[
Before
Front
State Lemma 5.20 about division in integral domains: (The quotient has what property?)
Back
State Lemma 5.20 about division in integral domains: (The quotient has what property?)
Lemma 5.20: In an integral domain, if \(a | b\) (i.e., \(b = ac\) for some \(c\)), then \(c\) is unique and is denoted by \(c = b/a\) (the quotient).
Explanation: If \(b = ac_1\) and \(b = ac_2\), then \(a(c_1 - c_2) = 0\). Since \(a \neq 0\) in an integral domain, we must have \(c_1 - c_2 = 0\)\(\implies c_1 = c_2\).
After
Front
State Lemma 5.20 about division in integral domains: (The quotient has what property?)
Back
State Lemma 5.20 about division in integral domains: (The quotient has what property?)
Lemma 5.20: In an integral domain, if \(a \mid b\) (i.e., \(b = ac\) for some \(c\)), then \(c\) is unique and is denoted by \(c = b/a\) (the quotient).
Explanation: If \(b = ac_1\) and \(b = ac_2\), then \(a(c_1 - c_2) = 0\). Since \(a \neq 0\) in an integral domain, we must have \(c_1 - c_2 = 0\)\(\implies c_1 = c_2\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Back | <p><strong>Lemma 5.20</strong>: In an integral domain, if \(a |
<p><strong>Lemma 5.20</strong>: In an integral domain, if \(a \mid b\) (i.e., \(b = ac\) for some \(c\)), then \(c\) is <strong>unique</strong> and is denoted by \(c = b/a\) (the quotient).</p> <p><strong>Explanation</strong>: If \(b = ac_1\) and \(b = ac_2\), then \(a(c_1 - c_2) = 0\). Since \(a \neq 0\) in an integral domain, we must have \(c_1 - c_2 = 0\)\(\implies c_1 = c_2\).</p> |