A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).
Note 1: ETH::DiskMat
Note Type: Horvath Cloze
GUID:
c.BJE1FC)A
Before
Front
Back
A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).
After
Front
A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).
Back
A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <p>A binary operation \(*\) on a set \(S\) is {{c1::associative}} if {{c2::\(a * (b * c) = (a * b) * c\)}} for all \( |
<p>A binary operation \(*\) on a set \(S\) is {{c1::associative}} if {{c2::\(a * (b * c) = (a * b) * c\)}} for all \(a, b, c\) in \(S\).</p> |
Note 2: ETH::DiskMat
Note Type: Horvath Cloze
GUID:
grVf##]DMH
Before
Front
Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).
Back
Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).
After
Front
Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).
Back
Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).
- \(a = a \cdot 1\)
- \(a = a \cdot 0\) (by assumption)
- \(a = 0\)
- Thus there is only the zero element, which is a contradiction to the non-triviality.
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Extra | Proof: Assume \(1 = 0\) for contradiction. For any \(a \in R\)<br><ol><li>\(a = a \cdot 1\)</li><li>\(a = a \cdot 0\) (by assumption)</li><li>\(a = 0\)</li><li>Thus there is only the zero element, which is a contradiction to the non-triviality.</li></ol> |
Note 3: ETH::DiskMat
Note Type: Horvath Cloze
GUID:
oPaK;$.R2B
Before
Front
An irreducible polynomial of degree \(\geq 2\) has no roots in the field.
Proof: If it had a root \(\alpha\), then \((x - \alpha)\) would divide it by Lemma 5.29, contradicting irreducibility.
Back
An irreducible polynomial of degree \(\geq 2\) has no roots in the field.
Proof: If it had a root \(\alpha\), then \((x - \alpha)\) would divide it by Lemma 5.29, contradicting irreducibility.
After
Front
An irreducible polynomial has no roots in the field. It has to have degree \(\geq 2\).
Back
An irreducible polynomial has no roots in the field. It has to have degree \(\geq 2\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <p>An {{c1::irreducible}} polynomial |
<p>An {{c1::irreducible}} polynomial has {{c3::no roots}} in the field. It has to have {{c2::degree \(\geq 2\)}}.</p> |
| Extra | <strong>Proof</strong>: If it had a root \(\alpha\), then \((x - \alpha)\) would divide it by Lemma 5.29, contradicting irreducibility. |
Note 4: ETH::DiskMat
Note Type: Horvath Cloze
GUID:
s`dg+)$g
Before
Front
If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).
Back
If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).
After
Front
If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).
Back
If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <p>If both {{c1::\(b * a = e\)}} and {{c2::\(a * b = e\)}}, then \( |
<p>If both {{c1::\(b * a = e\)}} and {{c2::\(a * b = e\)}}, then \(b\) is {{c4::simply called an inverse of \(a\)}}.</p> |