Anki Deck Changes

Commit: bfc6bd6a - fixes dm

Author: obrhubr <obrhubr@gmail.com>

Date: 2025-12-26T15:52:02+01:00

Changes: 4 note(s) changed (0 added, 4 modified, 0 deleted)

Note 1: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: c.BJE1FC)A
modified

Before

Front

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::2._Associativity_and_Monoids

A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).

Back

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::2._Associativity_and_Monoids

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

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::2._Associativity_and_Monoids

A binary operation \(*\) on a set \(S\) is associative if \(a * (b * c) = (a * b) * c\) for all \(a, b, c\) in \(S\).

Back

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::2._Associativity_and_Monoids

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 \({{c3::a, b, c}}\) in \(S\).</p> <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>
Tags: ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::2._Associativity_and_Monoids

Note 2: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: grVf##]DMH
modified

Before

Front

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::1._Definition_of_a_Ring

Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).

Back

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::1._Definition_of_a_Ring

Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).

After

Front

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::1._Definition_of_a_Ring

Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).

Back

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::1._Definition_of_a_Ring

Lemma 5.17(4): If a ring \(R\) is non-trivial (has more than one element), then \(1 \neq 0\).


Proof: Assume \(1 = 0\) for contradiction. For any \(a \in R\)
  1. \(a = a \cdot 1\)
  2. \(a = a \cdot 0\) (by assumption)
  3. \(a = 0\)
  4. 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&nbsp;\(1 = 0\)&nbsp;for contradiction. For any&nbsp;\(a \in R\)<br><ol><li>\(a = a \cdot 1\)</li><li>\(a = a \cdot 0\)&nbsp;(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>
Tags: ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::1._Definition_of_a_Ring

Note 3: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: oPaK;$.R2B
modified

Before

Front

ETH::1._Semester::DiskMat::5._Algebra::7._Polynomials_as_Functions::2._Roots

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

ETH::1._Semester::DiskMat::5._Algebra::7._Polynomials_as_Functions::2._Roots

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

ETH::1._Semester::DiskMat::5._Algebra::7._Polynomials_as_Functions::2._Roots

An irreducible polynomial has no roots in the field. It has to have degree \(\geq 2\).

Back

ETH::1._Semester::DiskMat::5._Algebra::7._Polynomials_as_Functions::2._Roots

An irreducible polynomial has no roots in the field. It has to have degree \(\geq 2\).


Proof: If it had a root \(\alpha\), then \((x - \alpha)\) would divide it by Lemma 5.29, contradicting irreducibility.
Field-by-field Comparison
Field Before After
Text <p>An {{c1::irreducible}} polynomial of degree {{c2::\(\geq 2\)}} has {{c3::no roots}} in the field.</p> <p><strong>Proof</strong>: If it had a root \(\alpha\), then \((x - \alpha)\) would divide it by Lemma 5.29, contradicting irreducibility.</p> <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&nbsp;\(\alpha\), then&nbsp;\((x - \alpha)\)&nbsp;would divide it by Lemma 5.29, contradicting irreducibility.
Tags: ETH::1._Semester::DiskMat::5._Algebra::7._Polynomials_as_Functions::2._Roots

Note 4: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: s`dg
modified

Before

Front

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups

If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).

Back

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups

If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).

After

Front

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups

If both \(b * a = e\) and \(a * b = e\), then \(b\) is simply called an inverse of \(a\).

Back

ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups

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 \({{c3::b}}\) is simply called an {{c4::inverse}} of \(a\).</p> <p>If both {{c1::\(b * a = e\)}} and {{c2::\(a * b = e\)}}, then \(b\) is {{c4::simply called an inverse of \(a\)}}.</p>
Tags: ETH::1._Semester::DiskMat::5._Algebra::2._Monoids_and_Groups::3._Inverses_and_Groups
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