Anki Deck Changes

Commit: 7c0cb589 - fix cloze issues

Author: obrhubr <obrhubr@gmail.com>

Date: 2025-12-18T06:40:40+01:00

Changes: 3 note(s) changed (0 added, 1 modified, 2 deleted)

Note 1: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: om)==wk?k1
modified

Before

Front

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::4._Zerodivisors_and_Integral_Domains

An integral domain \(D\) is a (nontrivial, \(0 \neq 1\)) commutative ring without c3:

Back

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::4._Zerodivisors_and_Integral_Domains

An integral domain \(D\) is a (nontrivial, \(0 \neq 1\)) commutative ring without c3:

After

Front

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::4._Zerodivisors_and_Integral_Domains

An integral domain \(D\) is a (nontrivial, \(0 \neq 1\)) commutative ring without zerodivisors (\(ab = 0 \implies a = 0 \lor b = 0\))

Back

ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::4._Zerodivisors_and_Integral_Domains

An integral domain \(D\) is a (nontrivial, \(0 \neq 1\)) commutative ring without zerodivisors (\(ab = 0 \implies a = 0 \lor b = 0\))

Field-by-field Comparison
Field Before After
Text <p>An {{c1::integral domain \(D\)}} is a {{c2::(nontrivial, \(0 \neq 1\)) commutative ring without c3::zerodivisors (\(ab = 0 \implies a = 0 \lor b = 0\))}}:</p> <p>An {{c1::integral domain \(D\)}} is a {{c2::(nontrivial, \(0 \neq 1\)) commutative ring}} without {{c3::zerodivisors (\(ab = 0 \implies a = 0 \lor b = 0\))}}</p>
Tags: ETH::1._Semester::DiskMat::5._Algebra::5._Rings_and_Fields::4._Zerodivisors_and_Integral_Domains

Note 2: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: sxW-Trt$`+
deleted

Deleted Note

Front

ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function

\(\mathbb{Z}_m^*\) is defined as?

Back

ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function

\(\mathbb{Z}_m^*\) is defined as?


\[ \overset{\text{def}}{=} \ \{a \in \mathbb{Z}_m \ | \ \gcd(a, m) = 1\} \]

This is the set of all elements in \(\mathbb{Z}_m\) that are coprime to \(m\).

Current

Note has been deleted

Field-by-field Comparison
Field Before After
Text <p>\(\mathbb{Z}_m^*\) is defined as?</p>
Extra <p>\[ \overset{\text{def}}{=} \ \{a \in \mathbb{Z}_m \ | \ \gcd(a, m) = 1\} \]</p> <p>This is the set of all elements in \(\mathbb{Z}_m\) that are coprime to \(m\).</p>
Tags: ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function

Note 3: ETH::DiskMat

Deck: ETH::DiskMat
Note Type: Horvath Cloze
GUID: G&Y|dtr7^k
deleted

Deleted Note

Front

ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function

If the prime factorization of \(m\) is \(m = \prod_{i=1}^r p_i^{e_i}\) then \(\varphi(m)\) is?

Back

ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function

If the prime factorization of \(m\) is \(m = \prod_{i=1}^r p_i^{e_i}\) then \(\varphi(m)\) is?


For a prime and  :\[\varphi(m) = \prod_{i=1}^r (p_i - 1)p_i^{e_i - 1}\]

Current

Note has been deleted

Field-by-field Comparison
Field Before After
Text <p>If the prime factorization of \(m\) is \(m = \prod_{i=1}^r p_i^{e_i}\) then \(\varphi(m)\) is?</p>
Extra <p>For a prime&nbsp;and&nbsp;&nbsp;:\[\varphi(m) = \prod_{i=1}^r (p_i - 1)p_i^{e_i - 1}\]<br></p>
Tags: ETH::1._Semester::DiskMat::5._Algebra::3._The_Structure_of_Groups::8._The_Group_Zₘ*_and_Euler's_Function
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