Real Number and Functions

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Real Number and Functions


    1. a)  A set of real numbers having a lower bound,

    2. b)  A set of real numbers without any lower bound,

    3. c)  A set of real numbers whose g.l.b. does not belong to it,

    4. d)  A bounded set of real numbers.

Answer:- (a) The set of Natural numbers (1,2, 3, ...) has a lower bound '0'.


Lower Bound:- A value that is less than or equal to every element of a set of data.

Example: in the set {2,4,6,8,10} 2 is a lower bound.

1 is also a lower bound (it is less than any element of that set), in fact any value 2 or less is a lower bound.

Answer:- (b) The set of Integers (.....0,1,2, 3, ...) has no lower bound.


Lower Bound:- A value that is less than or equal to every element of a set of data.

Example:- In the above example we can see there is no real number exists which is less then or equal to every element of set of integers

Answer:- (c) The greatest lower bound of the set

\[ \left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},……,\frac{1}{n}\right) \]


Here we can see 0 is the lower bound of the set but it does not belong to the set S.

Lower Bound:- A value that is less than or equal to every element of a set of data.

Greatest Lower Bound (g.l.b):-Let S be a nonempty set of real numbers that has a lower bound. A number c is the called the greatest lower bound (or the infimum, denoted infS) for S iff it satisfies the following properties:

1. c<=x for all x in S.

2. For all real numbers k, if k is a lower bound for S, then k<=c.

Answer:- (d) A closed interval [2,6] is a bounded set.


As it contains infinitely many real numbers but each number of this set is bounded below by 2 and bounded above by 6.

Bounded Set:- A set which is bounded below as well as bounded above.

Answer:- Consider a set S

\[ \left\{x\ \in\ R\ :\ x>0\right\} \]

Any real number p belongs to S so p > 0. i.e. 0 is a lower bound for the set S of positive real numbers. Thus the set S is bounded below. And Its infimum is 0.


The infimum is the greatest lower bound of a set S, defined as a quantity m such that no member of the set is less than m, but if epsilon is any positive quantity, however small, there is always one member that is less than m+epsilon. When it exists (which is not required by this definition, e.g., infR does not exist), the infimum is denoted infS or inf_(x in S)x.

Another Example:

The set S in above question has infimum 0, since
i) 0 c x for all x belongs to S and
ii) for each  epsilon>0 there is  epsilon/2 e S such that  epsilon/ 2 < O +  epsilon =  epsilon

\begin{array}{ l }\left(a\right)\ x=0\Leftrightarrow|x|=c\\\left(b\right)\ |xy|=|x|\cdot|y|\\\left(c\right)\ |1/x|=1/|x|,\quad\text{ if }x\ne0\\\left(d\right)\ |x-y|\le|x|+|y|\\\left(e\right)\ |x+y+z|\le|x|+|y|+|z|\\\left(f\right)\ |xyz|=|x|\cdot|y|\cdot|z|\end{array}


\( \begin{array}{l}\text{a) Forward Part:- By definition of Absolute value function}\\\begin{array}{l}\left|x\right|=\max\left\{x,-x\right\}\\if\ x=0\ then\ \left|x\right|=\left\{0,-0\right\}\\\left|x\right|=0\\\text{So if x = 0 then}\ \left|x\right|=0.\\\text{Now we will prove Converse Part:}\\if\ \left|x\right|=0\\\max\left\{x,-x\right\}=0\\then\ x=0\\\text{Hence Proved}\end{array}\end{array} \)

\( \begin{array}{l}\left(b\right)\ |xy|=\sqrt{(xy)^2}\\=\sqrt{x^2\cdot y^2}\\=\sqrt{x^2}\cdot\sqrt{y^2}\\=|x||y|\end{array} \)

\( \ \begin{array}{ c }\text{c) There are two cases}\text{: Case (a) if }x>0,|x|=x\text{ and }|1/x|=1/x=1/|x|\\\text{Case (b) }\text{if }x<0,|x|=-x\text{ and }|1/x|=-1/x=1/|x|\end{array} \)

\( \begin{array} { l } { \left. \begin{array}{l}d){ | x - y | = | x + ( - y ) | }\\{ \leq | x | + | - y | = | x | + | y | [ \because | - y | = | y | ] }\end{array} \right. } \\ { | x - y | \leq | x | + | y | } \end{array} \)

\( \begin{array}{ l }\text{ e) }|x+y+z|=|(x+y)+z|\\\le|x+y|+|z|\\\le|x|+|y|+|z|\\\\f)\quad|xyz|=|xy|z|\\\ \ \ \ \ \ \ \ \ \ \ =|x||y||z|\end{array} \)

\( \begin{array}{l}\int \:e^x\cos \left(x\right)dx\\=e^x\sin \left(x\right)-\int \:e^x\sin \left(x\right)dx\\=e^x\sin \left(x\right)-\left(-e^x\cos \left(x\right)-\int \:-e^x\cos \left(x\right)dx\right)\\=e^x\sin \left(x\right)-\left(-e^x\cos \left(x\right)-\left(-\int \:e^x\cos \left(x\right)dx\right)\right)\\\int \:e^x\cos \left(x\right)dx=e^x\sin \left(x\right)-\left(-e^x\cos \left(x\right)-\left(-\int \:e^x\cos \left(x\right)dx\right)\right)\\=\frac{e^x\sin \left(x\right)}{2}+\frac{e^x\cos \left(x\right)}{2}+C\end{array} \)

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