Some real analysis exercises on:
– Triangle Inequality
– Archimedean Property
– Limit of a sequence is unique and convergence sequence is bounded
– Convergence sequence
Triangle Inequality
$$ { \text{ Prove } |a + b| \leq |a| + |b| }$$Proof
$$ { \text{ Show, } \left( |a| + |b| \right)^2 \geq |a + b|^2 \text{ noting, } \forall x \in \mathbb{R} \text{ , } |x|^2 = x^2 }$$$${ \forall a,b \in \mathbb{R} \text{ , } \left( |a| + |b| \right)^2 }$$
$${ = |a|^2 + 2.|a|.|b| + |b|^2 }$$ $${ \geq a^2 + 2.a.b + b^2 }$$ $${ = (a + b)^2 \Rightarrow \left|a + b\right|^2 }$$ $${ \text{ So, } \left( |a| + |b| \right)^2 \geq |a + b|^2 }$$ $${ \text{ So, } |a| + |b| \geq |a + b| }$$ $${ |a + b| \leq |a| + |b| \text{ ∎ } }$$Archimedean Property
$$ { \text{For any number c, there exists a natural number n that is greater than c.} }$$ $$ { \text{ Prove that for every positive number } \varepsilon , } $$ $$ { \text{ there exists a positive integer N such that } \frac {1}{n} \lt \varepsilon \text{ for all n } \geq N .}$$Proof
$$ { \text{ Let, } \varepsilon > 0 \text{ be any positive number. Note } \frac {1}{\varepsilon} > 0, } $$ $$ { \text{ So by the A.P.} , \exists N \in \mathbb{N} \text{ s.t } N > \frac {1}{\varepsilon}.\textit{ *Note N derivation is left to reader } } $$ $$ { \text{ Thus } \varepsilon.N >1 , \text{ so } \varepsilon > \frac {1}{N}. \text{ then} \forall n \geq N, } $$ $$ { \frac {1}{N} \geq \frac {1}{n}; \text{ that is } \frac {1}{n} \leq \frac {1}{N} < \varepsilon \text{ ∎ } } $$Limit of a sequence is unique
$$ { \text{Prove that a sequence is unique, recall } a_n \to L \text{ , that is } \forall \varepsilon >0 \exists N \in \mathbb{Z}^+ s.t. \forall n > N |a_n - L| < \varepsilon } $$Proof
$$ { \text{ Spse. } a_n \to a \text{ and } a_n \to b \text{ with } a \neq b. } $$ $$ { \text{ Put } \varepsilon = |a - b| >0 } $$ $$ { \text{ Since } a_n \to a, \exists N \in \mathbb{Z}^+ s.t. \forall n > N_1, |a_n - a| < \frac {\varepsilon}{2}.} $$ $$ { \text{ Since } a_n \to b, \exists N \in \mathbb{Z}^+ s.t. \forall n > N_2, |a_n - b| < \frac {\varepsilon}{2}. }$$ $$ { \text{ Set M = max} (N_1 ,N_2) \text{ Then} \forall n > M,|a - b| = |a - a_n+a_n -b| }$$ $$ { \leq |a - a_n| + |a_n -b| }$$ $$ { \le \frac {\varepsilon}{2} +\frac {\varepsilon}{2} = \varepsilon = |a - b|. }$$ $$ { \text{ Thus } |a - b| < |a - b|. \text{ a contradiction} \Rightarrow \Leftarrow }$$Convergence proof
$$ { \text{Prove } \lim_{n\to\infty} \frac {n}{n+1} =1 } $$ $$ { \text{Recall } \lim_{n\to\infty} a_n = L \text{ means } \forall \varepsilon > 0 N \exists N \in \mathbb{Z}^+ \text{ s.t , } \forall n > N , |a_n - L| < \varepsilon } $$Proof
$$ { \text{ Let, } \varepsilon > 0 \text{ . Choose } N > \frac {1}{\varepsilon} > 0, } $$ $$ { \text{ So by the A.P.} , \exists N \in \mathbb{N} \text{ s.t } N > \frac {1}{\varepsilon}.\textit{ *Note N derivation is left to reader } } $$ $$ { \text{ Then } \forall n > N\varepsilon.N >1 , |\frac {n}{n+1} -1 | = |\frac {n}{n+1} - \frac {n+1}{n+1} | } $$ $$ { = |\frac {n - (n+1)}{n+1} | = | \frac {-1}{n+1} | = \frac {1}{n+1} \leq \frac {1}{n} } $$ $$ { \text{ Since } n > N > \frac {1}{\varepsilon} \text{ therefore } \frac {\varepsilon.n}{n} > \frac {1}{n} } $$ $$ { \varepsilon > \frac {1}{n} \text{ So, } \frac {1}{n} < \varepsilon } $$ $$ { \text{ Thus, } | \frac {n}{n+1} - 1 | \leq \frac {1}{n} < \varepsilon. ∎ } $$