Learn Pure Mathematics

Some important point only for real numbers

1- Denominator of any real number can not be zero.

Y= N/D D≠0

Hence if denominator is zero then real number can never be possible.

So 1/0, 2/0, 3/0,………..are practically impossible.

i.e. 1/0= undefined.

Theorem:- Let a be any real number and b any +ve real number. Then there exists a positive integer n such that

nb>a

Proof:- Given that aϵ R & bϵ R⁺

So there are two possible cases.

Case 1:- When a≤0

In this case the relation nb>a is always true because the value of nb is always +ve.

Case 2:- When a>0

Let us that there exists no +ve integar such that nb>a

Then we have nb≤a ⩝ n∈N

It means that a be the upper bound of set S which is given by

S= {b,2b,3b,………..} = {nb:n∈N}.

Upper bound of a subset of R:-

Let S be a subset of real numbers. If there exist a real number K , such that x≤K ⩝ x∈S

Then K is called an upper bound of set S.

If there exists an upper bound for a set S then it is called “bounded above”.

Example:- The set S= {………-4,-3,-2,-1} is bounded above & 9 is an upper bound.