Question

Proving for $a>1,b > 1$, $2^a \ne p^b + 1$

score 82 · Answer 1 · 2026-04-01 02:56:52

82

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Proving for $a>1,b > 1$, $2^a \ne p^b + 1$

Published on 01 Apr 2026 - 2:56

score 487 · Answer 2 · 2026-03-26 15:17:17

487

Views

Prove that $1+x<e^x$ for all real numbers $x\neq0$.

Published on 26 Mar 2026 - 15:17

score 70 · Answer 3 · 2026-04-01 00:03:50

70

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Where is the error in this proof?

Published on 01 Apr 2026 - 0:03

score 68 · Answer 4 · 2018-08-28 07:58:16

68

Views

Cohn, Exercise 2.4.11

Published on 28 Aug 2018 - 7:58

score 308 · Answer 5 · 2026-03-25 17:24:55

308

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$A^3 = A^2$ How can $A$'s minimal polynomial look like?

Published on 25 Mar 2026 - 17:24

score 28 · Answer 6 · 2018-08-28 14:49:16

28

Views

Question regarding the formulation of a proof using the mean value theorem for integrals

Published on 28 Aug 2018 - 14:49

score 74 · Answer 7 · 2026-03-30 15:29:59

74

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$U_1 \not\subseteq U_2$, $U_2 \not\subseteq U_1$ $\implies$ $U_1 \cup U_2$ is not a subspace

Published on 30 Mar 2026 - 15:29

score 131 · Answer 8 · 2026-03-25 07:53:26

131

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If $\mathbb{P}$ is an infinite notion of forcing then there is $H\subset \mathbb{P}$ such that $M[H]$ is not a model of ZFC.

Published on 25 Mar 2026 - 7:53

score 616 · Answer 9 · 2018-08-28 18:36:51

616

Views

Entry Level Combinatorics Proof-Check

Published on 28 Aug 2018 - 18:36

score 180 · Answer 10 · 2026-04-01 09:59:45

180

Views

Cohn, Exercise 2.4.11, Intuition behind my solution?

Published on 01 Apr 2026 - 9:59

score 549 · Answer 11 · 2026-04-03 23:22:25

549

Views

Homotopy type of the complement of a subspace

Published on 03 Apr 2026 - 23:22

score 235 · Answer 12 · 2026-04-03 23:24:11

235

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Exercise about retract of $GL(n, \mathbb{C})$ homeomorphic to $S^1$

Published on 03 Apr 2026 - 23:24

score 68 · Answer 13 · 2026-04-01 03:33:41

68

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Proving that $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}|e^{z^2}| = e^{(\Re z)^2-(\Im z)^2}$

Published on 01 Apr 2026 - 3:33

score 352 · Answer 14 · 2026-04-01 03:36:38

352

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Proving $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}\Re{(\cosh z}) = \cosh(\Re z)\cdot \cos(\Im z)$

Published on 01 Apr 2026 - 3:36

score 6.5k · Answer 15 · 2026-03-26 08:00:31

6.5k

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$H$ is normal subgroup of $G$ iff $gHg^{-1}=H$ for every $g$ of $G$

Published on 26 Mar 2026 - 8:00

List Question

Proving for $a>1,b > 1$, $2^a \ne p^b + 1$

Prove that $1+x<e^x$ for all real numbers $x\neq0$.

Where is the error in this proof?

Cohn, Exercise 2.4.11

$A^3 = A^2$ How can $A$'s minimal polynomial look like?

Question regarding the formulation of a proof using the mean value theorem for integrals

$U_1 \not\subseteq U_2$, $U_2 \not\subseteq U_1$ $\implies$ $U_1 \cup U_2$ is not a subspace

If $\mathbb{P}$ is an infinite notion of forcing then there is $H\subset \mathbb{P}$ such that $M[H]$ is not a model of ZFC.

Entry Level Combinatorics Proof-Check

Cohn, Exercise 2.4.11, Intuition behind my solution?

Homotopy type of the complement of a subspace

Exercise about retract of $GL(n, \mathbb{C})$ homeomorphic to $S^1$

Proving that $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}|e^{z^2}| = e^{(\Re z)^2-(\Im z)^2}$

Proving $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}\Re{(\cosh z}) = \cosh(\Re z)\cdot \cos(\Im z)$

$H$ is normal subgroup of $G$ iff $gHg^{-1}=H$ for every $g$ of $G$

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