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seen Jun 16 at 8:56

Dec
11
comment Necessity for the equality of two subspaces
What does "without using the other inclusion" exactly mean?
Dec
10
awarded  Editor
Dec
10
revised If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $r$ such that $x<r<y$
added 102 characters in body
Dec
10
comment If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $r$ such that $x<r<y$
For negatives you should say something. If $x$ and $y$ are negative, multiply by $-n$, etc. Otherwise $y$ is positive and we need to find $n$ and $m$, such that $0 < m \leq ny$. Should I remove my "answer"?
Dec
10
comment If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $r$ such that $x<r<y$
Let $m$ be the smallest natural number, which satisfies $nx < m$ (exists because of well ordering). Then $m - xn <= 1$ because otherwise is $xn < m-1$.
Dec
10
answered If $x\in \mathbb{R}$,$y\in \mathbb{R}$ and $x<y$,then there exists a rational $r$ such that $x<r<y$
Dec
9
comment A simple limit of a sequence
Why not $1 < \sqrt[n]{1 + (1/n)} < 1 + (1/n)$?
Dec
9
awarded  Supporter
Dec
9
answered How to find $\mathrm{Rank}(A)$
Dec
9
answered Continuous functions proof
Dec
8
awarded  Analytical
Dec
8
awarded  Teacher
Dec
8
awarded  Custodian