Given:

g(x):R---->R is continuous

g(0)=1

Show that if there exists a +ve no. 'a' where g is zero..i.e. g(a)=0, then there exists a smallest +ve no. 'k' where g(k)=0.

Hint: Consider the set of all positive x where g(x)=0

 

Note: This is the complete question,Concepts which may be used: Convergence of sequences and Continuity of a function.

We will prove it by contradiction . We will assume that there does not exist smallest positive number ?k? such that ?g(k)=0? .

Then we will construct a sequence ?{xn?}? converges to ?0? such that ?g(xn?)=0? for all ?n∈N? .

Then using sequential criterion of continuity we will prove that ?g(0)=0? which will be a contradiction to ?g(0)=1? .

Then we can conclude that there exist smallest positive number ?k? such that ?g(k)=0? .

Step-by-step explanation

?g:R→R? be a continuous function such that ?g(0)=1? and there exist ?a>0? such that ?g(a)=0? . We will prove by contradiction that there exist smallest positive number ?k? such that ?g(k)=0? .

Suppose there does not exist smallest positive number ?k? such that ?g(k)=0? then for all ?r>0? there exist ?k<r? such that ?g(k)=0? .

For all ?n∈N? if we choose ?r=n1?>0? then there exist ?0<? ?k<r? such that ?g(k)=0? call it ?xn?? that is ?g(xn?)=0? .

So for all ?n∈N? , ?0<xn?<n1??

Taking limit both side we get ,

?0≤limn→∞?xn?≤limn→∞?n1??

?⇒0≤limn→∞?xn?≤0?

?⇒limn→∞?xn?=0? , by sandwich lemma .

As ?g(xn?)=0? for all ?n∈N? so the sequence ?{g(xn?)}? converges to ?0? .

As ?{xn?}? converges to ?0? and ?g? is continuous so the sequence ?{g(xn?)}? converges to ?g(0)? . But since ?{g(xn?)}? converges to ?0? so by uniqueness of limit ?g(0)=0? , a contradiction to ?g(0)=1? .

Hence there exist smallest positive number ?k? such that ?g(k)=0? .