Real Number and Functions
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Real Number and Functions


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

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

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

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.
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 , 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\Leftrightarrowx=c\\\left(b\right)\ xy=x\cdoty\\\left(c\right)\ 1/x=1/x,\quad\text{ if }x\ne0\\\left(d\right)\ xy\lex+y\\\left(e\right)\ x+y+z\lex+y+z\\\left(f\right)\ xyz=x\cdoty\cdotz\end{array}
Solution:
\( \begin{array}{l}\text{a) Forward Part: By definition of Absolute value function}\\\begin{array}{l}\leftx\right=\max\left\{x,x\right\}\\if\ x=0\ then\ \leftx\right=\left\{0,0\right\}\\\leftx\right=0\\\text{So if x = 0 then}\ \leftx\right=0.\\\text{Now we will prove Converse Part:}\\if\ \leftx\right=0\\\max\left\{x,x\right\}=0\\then\ x=0\\\text{Hence Proved}\end{array}\end{array} \)
\( \begin{array}{l}\left(b\right)\ xy=\sqrt{(xy)^2}\\=\sqrt{x^2\cdot y^2}\\=\sqrt{x^2}\cdot\sqrt{y^2}\\=xy\end{array} \)
\( \ \begin{array}{ c }\text{c) There are two cases}\text{: Case (a) if }x>0,x=x\text{ and }1/x=1/x=1/x\\\text{Case (b) }\text{if }x<0,x=x\text{ and }1/x=1/x=1/x\end{array} \)
\( \begin{array} { l } { \left. \begin{array}{l}d){  x  y  =  x + (  y )  }\\{ \leq  x  +   y  =  x  +  y  [ \because   y  =  y  ] }\end{array} \right. } \\ {  x  y  \leq  x  +  y  } \end{array} \)
\( \begin{array}{ l }\text{ e) }x+y+z=(x+y)+z\\\lex+y+z\\\lex+y+z\\\\f)\quadxyz=xyz\\\ \ \ \ \ \ \ \ \ \ \ =xyz\end{array} \)
\( \begin{array}{l}\int \:e^x\cos \left(x\right)dx\\=e^x\sin \left(x\right)\int \:e^x\sin \left(x\right)dx\\=e^x\sin \left(x\right)\left(e^x\cos \left(x\right)\int \:e^x\cos \left(x\right)dx\right)\\=e^x\sin \left(x\right)\left(e^x\cos \left(x\right)\left(\int \:e^x\cos \left(x\right)dx\right)\right)\\\int \:e^x\cos \left(x\right)dx=e^x\sin \left(x\right)\left(e^x\cos \left(x\right)\left(\int \:e^x\cos \left(x\right)dx\right)\right)\\=\frac{e^x\sin \left(x\right)}{2}+\frac{e^x\cos \left(x\right)}{2}+C\end{array} \)