Definition: Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses. It can be defined for various spaces, such as:
For real numbers, a sequence $\langle x_n\rangle$ of real numbers, such that:
$\qquad\quad\langle x_n\rangle = x_1, x_2, x_3, \cdots$ where, $x_1, x_2, x_3, \cdots\;\in\;\mathbb{R}$
is a Cauchy sequence, if for every positive real number $\epsilon$, $\exists$ an $N\;\in\;\mathbb{N}$, such that
$\qquad\quad |x_n - x_m| < \epsilon \;\;\forall\;\; n, m\;\geq\; N$
Similarly, for rational numbers, a sequence $\langle x_n\rangle$ of rational numbers, such that:
$\qquad\quad\langle x_n\rangle = x_1, x_2, x_3, \cdots$ where, $x_1, x_2, x_3, \cdots\;\in\;\mathbb{Q}$
is a Cauchy sequence, if for every positive rational number $\epsilon$, $\exists$ an $N\;\in\;\mathbb{N}$, such that
$\qquad\quad |x_n - x_m| < \epsilon \;\;\forall\;\; n, m\;\geq\; N$
For example:
$\textrm{If}\;\;\;\quad \langle x_n\rangle =\left\{\dfrac{1}{n}\;\big|\;\; n\;\in\;\mathbb{Z}\right\}\\
\begin{align}\textrm{Then,}\; |x_n - x_m| &= \left| \dfrac{1}{n} - \dfrac{1}{m}\right|\\[6pt]
& < \dfrac{1}{n} \text{&} \dfrac{1}{m}\end{align}$
Choosing:
$\begin{align}\qquad\quad & N > \dfrac{1}{\epsilon}\\
\Rightarrow\quad\;\; &\dfrac{1}{N} < \epsilon\\
\therefore\quad\;\;\; &|x_n - x_m|\;\;< \; \dfrac{1}{n} \text{&} \dfrac{1}{m} < \;\epsilon\;\;\forall\;\; n, m\;\geq\; N \;\left(\textrm{for}\; N > \dfrac{1}{\epsilon}\right) \end{align}$
Hence, $\langle x_n\rangle =\left\{\dfrac{1}{n}\;\big|\;\; n\;\in\;\mathbb{Z}\right\}$ is a Cauchy sequence.
Abelian Group
Definition: An Abelian group, or a commutative group, is a group which along with the group axioms, satisfies an additional axiom of commutativity, i.e., if $(G, •)$ is an abelian group, then it satisfies the following five axioms:
Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.
Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.
Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$
$\qquad\quad e • a = a • e = a$
Such $e$ is known as the identity element of $G$ w.r.t. $•$.
Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,
$\qquad\quad a • b = b • a = e$
Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.
Commutativity:
$\forall\;\;a, b\;\in\; G$,
$\qquad\quad a • b = b • a$
Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.
Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.
Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$
$\qquad\quad e • a = a • e = a$
Such $e$ is known as the identity element of $G$ w.r.t. $•$.
Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,
$\qquad\quad a • b = b • a = e$
Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.
Commutativity:
$\forall\;\;a, b\;\in\; G$,
$\qquad\quad a • b = b • a$
Group
Definition: A group is a set, $G$, together with an operation $•$ that combines any two elements $a$ and b to form another element, denoted by $a • b$ or $ab$. A group $(G, •)$, must satisfy four requirements known as the group axioms:
Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.
Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.
Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$
$\qquad\quad e • a = a • e = a$
Such $e$ is known as the identity element of $G$ w.r.t. $•$.
Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,
$\qquad\quad a • b = b • a = e$
Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.
Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.
Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.
Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$
$\qquad\quad e • a = a • e = a$
Such $e$ is known as the identity element of $G$ w.r.t. $•$.
Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,
$\qquad\quad a • b = b • a = e$
Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.
Rational Numbers
Definition: a rational number is any number that can be written in the p/q form of two integers, p and q, with the denominator q not equal to zero. The set of all rational numbers is denoted by $\mathbb{Q}$, i.e.,
$\qquad\quad\mathbb{Q} = \left\{ \dfrac{p}{q} \;\big|\;\; p, q\;\in\;\mathbb{Z};\;\; q\neq 0\right\}$
In decimal representation, a number is a rational number if and only if the decimal expansion of a number either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. This holds true irrespective of the base, i.e, decimal, binary, etc.
Related:
Integers (definition)
$\qquad\quad\mathbb{Q} = \left\{ \dfrac{p}{q} \;\big|\;\; p, q\;\in\;\mathbb{Z};\;\; q\neq 0\right\}$
In decimal representation, a number is a rational number if and only if the decimal expansion of a number either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. This holds true irrespective of the base, i.e, decimal, binary, etc.
Related:
Integers (definition)
Integers
Definition: An integer is a number that can be written without a fractionl component. The set of integers is denoted by $\mathbb{Z}$. Further, if $n$ is an integer, then $n^+$ (or the successor of $n$, eg. $0^+ = 1$, $1^+ = 2$) is also an integer. Similarly, if $n$ is an integer, then $n^-$ is also an integer.
Median Through Hypotenuse
Theorem: In a right triangle, length of a median drawn through the vertex having right angle to meet hypotenuse, is equal to one half of the length of the hypotenuse.
Prerequisites:
Median (definition)
Midpoint Theorem (proof)
SAS congruence (proof)
Angle on a straight line (proof)
Corresponding angles property (proof)
Proof:
Let $\triangle ABC$ be a right triangle, right angled at B. Let BD be a median drawn from B to meet AC at D.
We need to show that BD = $\dfrac{1}{2}$ AC. For this, let us join DE, where E is the midpoint of AB.
Since, D and E are the midpoints of AC and AB respectively (see definition of median), hence by Midpoint Theorem,
$\begin{align}\qquad\quad &DE \parallel BC\\
\Rightarrow\quad\;\; &\angle AED = \angle ABC = 90^o\quad\qquad\qquad\!\! && \text{(corresponding angles)} &&& \cdots\text{(1)}\\
\Rightarrow\quad\;\; & \angle BED = 180^o -\angle AED = 90^o && \text{(angle on a straight line)} &&& \cdots\text{(2)}\end{align}$
Now, consider $\triangle AED$ and $\triangle BED$:
$\begin{align}\qquad\quad\;\; & AE = BE\qquad\qquad\qquad\qquad\qquad && \text{(by construction)}\\
& DE = DE && \text{(common)}\\
& \angle AED = \angle BED = 90^o && \text{(from $(1)$ and $(2)$)}\\
\therefore\quad\;\;\;\;\; & \triangle AED\cong\triangle BED && \text{(by SAS congruence)}\\
\\[12pt]
\Rightarrow\;\quad\quad & AD = BD && \text{(CPCTC)}\qquad\qquad\qquad\qquad\qquad\cdots\text{(3)}\\
\text{But, }\;\quad & AD = AC/2 && \text{(by definition of median)}\\
\text{Hence, }\; & BD = AC/2 && \text{(from $(3)$)}\end{align}$
Q.E.D.
Prerequisites:
Median (definition)
Midpoint Theorem (proof)
SAS congruence (proof)
Angle on a straight line (proof)
Corresponding angles property (proof)
Proof:
Let $\triangle ABC$ be a right triangle, right angled at B. Let BD be a median drawn from B to meet AC at D.
We need to show that BD = $\dfrac{1}{2}$ AC. For this, let us join DE, where E is the midpoint of AB.
Since, D and E are the midpoints of AC and AB respectively (see definition of median), hence by Midpoint Theorem,
$\begin{align}\qquad\quad &DE \parallel BC\\
\Rightarrow\quad\;\; &\angle AED = \angle ABC = 90^o\quad\qquad\qquad\!\! && \text{(corresponding angles)} &&& \cdots\text{(1)}\\
\Rightarrow\quad\;\; & \angle BED = 180^o -\angle AED = 90^o && \text{(angle on a straight line)} &&& \cdots\text{(2)}\end{align}$
Now, consider $\triangle AED$ and $\triangle BED$:
$\begin{align}\qquad\quad\;\; & AE = BE\qquad\qquad\qquad\qquad\qquad && \text{(by construction)}\\
& DE = DE && \text{(common)}\\
& \angle AED = \angle BED = 90^o && \text{(from $(1)$ and $(2)$)}\\
\therefore\quad\;\;\;\;\; & \triangle AED\cong\triangle BED && \text{(by SAS congruence)}\\
\\[12pt]
\Rightarrow\;\quad\quad & AD = BD && \text{(CPCTC)}\qquad\qquad\qquad\qquad\qquad\cdots\text{(3)}\\
\text{But, }\;\quad & AD = AC/2 && \text{(by definition of median)}\\
\text{Hence, }\; & BD = AC/2 && \text{(from $(3)$)}\end{align}$
Q.E.D.
Centroid
Theorem: All the three medians of a triangle intersect at a single point called centroid, which divides each median in a ratio of $2:1$.
Prerequisites:
Similarity of triangles (definition)
Midpoint Theorem (proof)
AA similarity (proof)
Alternate angles property (proof)
vertical angle theorem (proof)
Proof:
Let ABC be a triangle having medians AD, BE and CF.
Let us consider the medians BE and CF, which intersect at a point G. Let us join E and F by a straight line.
Since EF joins the midpoints of the lines AB and AC, hence by midpoint theorem:
$\qquad\quad EF \parallel BC$
And, $\;\;\;\! EF = \dfrac{1}{2} BC\qquad\qquad\qquad\cdots\text{(1)}$
Now, consider triangles $\triangle BCG$ and $\triangle EFG$:
$\qquad\quad\angle EGF = \angle BGC\qquad\qquad\qquad\text{(vertically opposite angles)}\\
\qquad\quad\angle GFE = \angle GCB\qquad\qquad\qquad\text{(alternate angles)}\\
\therefore\quad\;\;\triangle BCG\sim\triangle EFG\qquad\qquad\qquad\text{(AA similarity)}$
Thus, by definition of similarity, corresponding sides are proportional, hence:
$\qquad\quad\dfrac{GE}{GB} = \dfrac{GF}{GC} =\dfrac{EF}{BC} = \dfrac{1}{2}\qquad\qquad\text{(from $(1)$)}$
Thus, G divides BE and CF in the ratio $2:1$.
Similarly, by considering the medians AD and BE, which intersect at a point G', it can be shown that G' divides AD and BE in the ratio 2:1.
But BE is divided in $2:1$ ratio by G. Hence, G' $=$ G.
Thus, all the medians intersect at a single point, which divides the medians in a ratio $2:1$.
Q.E.D.
Recommended;
Median through hypotenuse
Basic Proportionality Theorem
Angle Bisector Theorem
Prerequisites:
Similarity of triangles (definition)
Midpoint Theorem (proof)
AA similarity (proof)
Alternate angles property (proof)
vertical angle theorem (proof)
Proof:
Let ABC be a triangle having medians AD, BE and CF.
Let us consider the medians BE and CF, which intersect at a point G. Let us join E and F by a straight line.
Since EF joins the midpoints of the lines AB and AC, hence by midpoint theorem:
$\qquad\quad EF \parallel BC$
And, $\;\;\;\! EF = \dfrac{1}{2} BC\qquad\qquad\qquad\cdots\text{(1)}$
Now, consider triangles $\triangle BCG$ and $\triangle EFG$:
$\qquad\quad\angle EGF = \angle BGC\qquad\qquad\qquad\text{(vertically opposite angles)}\\
\qquad\quad\angle GFE = \angle GCB\qquad\qquad\qquad\text{(alternate angles)}\\
\therefore\quad\;\;\triangle BCG\sim\triangle EFG\qquad\qquad\qquad\text{(AA similarity)}$
Thus, by definition of similarity, corresponding sides are proportional, hence:
$\qquad\quad\dfrac{GE}{GB} = \dfrac{GF}{GC} =\dfrac{EF}{BC} = \dfrac{1}{2}\qquad\qquad\text{(from $(1)$)}$
Thus, G divides BE and CF in the ratio $2:1$.
Similarly, by considering the medians AD and BE, which intersect at a point G', it can be shown that G' divides AD and BE in the ratio 2:1.
But BE is divided in $2:1$ ratio by G. Hence, G' $=$ G.
Thus, all the medians intersect at a single point, which divides the medians in a ratio $2:1$.
Q.E.D.
Recommended;
Median through hypotenuse
Basic Proportionality Theorem
Angle Bisector Theorem
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