# Real Number and Functions

## 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'.

Explanation:-

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.

Explanation:-

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

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

Explanation:-

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  is the called the greatest lower bound (or the infimum, denoted ) for  iff it satisfies the following properties:

1.  for all .

2. For all real numbers , if  is a lower bound for , then .

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

Explanation:-

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.

$\left\{x\ \in\ R\ :\ x>0\right\}$

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 , defined as a quantity  such that no member of the set is less than , but if  is any positive quantity, however small, there is always one member that is less than . When it exists (which is not required by this definition, e.g.,  does not exist), the infimum is denoted  or .

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  >0 there is  /2 e S such that  / 2 < O +   =

\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}

Solution:-

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

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

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

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

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

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

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