any nonzero rational number can be expressed uniquely as the ratio p/q of two in
ID: 3077592 • Letter: A
Question
any nonzero rational number can be expressed uniquely as the ratio p/q of two integers p and q. where p and q have no common factors (other than +&- 1) and q>0. using this representation define a function f(x) for 0<x<1 byf(x)= { 1/q, x rational, x=p/q as above
{0, x irrational
thus, f(1/3)=f(2/3)=1/3, f(1/4)=f(3/4)=1/4, f(2/4)=1/2, and f(1/ square root of 2)=0
a) show that f is not continuous at any rational point c in (0,1)
b) show that f is continuous at any irrational point c in (0,1).
Explanation / Answer
any nonzero rational number can be expressed uniquely as the ratio p/q of two integers p and q. where p and q have no common factors (other than +&- 1) and q>0. using this representation define a function f(x) for 0<x<1 byf(x)= { 1/q, x rational, x=p/q as above
{0, x irrational
thus, f(1/3)=f(2/3)=1/3, f(1/4)=f(3/4)=1/4, f(2/4)=1/2, and f(1/ square root of 2)=0
a) show that f is not continuous at any rational point c in (0,1)
b) show that f is continuous at any irrational point c in (0,1). AT THE OUT SET LET US NOTE THAT IN THE NEIGHBOURHOOD AROUND ANY RATIONAL NUMBER THERE ARE INFINITE IRRATIONAL NUMBERS AND IN THE NEIGHBOURHOOD AROUND ANY IRRATIONAL NUMBER THERE ARE INFINITE RATIONAL NUMBERS a) show that f is not continuous at any rational point c in (0,1) SO NOW LET US TAKE ANY RATIONAL POINT A=P/Q IN THE GIVEN INTERVAL FOR THE FUNCTION TO BE CONTINUOUS HERE WE SHALL HAVE LIMIT OF F[X] AS X TENDS TO A=P/Q SHALL EXIST AND SHALL BE F[A].......................1 SINCE A IS RATIONAL , WE GET F[A]=F[P/Q]=1/Q NOW LET US CHECK THE LIMIT IS 1/Q OR NOT IT WILL BE THE LIMIT IF WE CAN MAKE |F[X]-[1/Q]|<E >0.................WHERE E IS A SPECIFIED SMALL NUMBER BY TAKING X AS CLOSE TO A=P/Q AS DESIRED BUT FROM THE STATEMENT MADE AT THE BEGINING , THERE ARE INFINITE IRRATIONAL POINTS AROUND EVERY RATIONAL POINT. SO F[X] AT THAT POINT = 0 AS PER HYPOTHESIS. SO WE GET |0-[1/Q]<E............IMPOSSIBLE SINCE Q IS A GIVEN CONSTANT NUMBER,AND SO WE CAN DO IT IN NO WAY BY TAKING X ANY CLOSER TO A SO F[X] FAS NO LIMIT AS X TENDS TO A=P/Q A RATIONAL NUMBER SO THE FUNCTION IS DISCONTINUOUS AT ANY RATIONAL NUMBER N THE GIVEN INTERVAL. b) show that f is continuous at any irrational point c in (0,1). SO NOW LET US TAKE ANY IRRATIONAL POINT A IN THE GIVEN INTERVAL FOR THE FUNCTION TO BE CONTINUOUS HERE WE SHALL HAVE LIMIT OF F[X] AS X TENDS TO A SHALL EXIST AND SHALL BE F[A].......................2 SINCE A IS IRRATIONAL , WE GET F[A]=0.......AS GIVEN NOW LET US CHECK THE LIMIT IS 0 OR NOT IT WILL BE THE LIMIT IF WE CAN MAKE |F[X]-0]|<E >0.................WHERE E IS A SPECIFIED SMALL NUMBER BY TAKING X AS CLOSE TO A AS DESIRED BUT FROM THE STATEMENT MADE AT THE BEGINING , THERE ARE INFINITE RATIONAL POINTS AROUND EVERY IRRATIONAL POINT. SO F[X] AT THAT POINT = P/Q = 1/Q AS PER HYPOTHESIS. SO WE GET ||[1/Q]-0|<E............POSSIBLE SINCE A= P/Q IS A VARIABLE NUMBER NEAR ZERO ...THAT IS Q COULD BE VERY LARGE WITH P=1 AND SO WE CAN DO IT BY TAKING X CLOSER TO A ....SAY BY MAKING Q=R>1/E, WHERE R IS AN INTEGER SO F[X] FAS LIMIT EQUAL TO ZERO AS X TENDS TO A AN IRRATIONAL NUMBER WHICH IS ALSO EQUAL TO F[A] SO THE FUNCTION IS CONTINUOUS AT ANY IRRATIONAL NUMBER IN THE GIVEN INTERVAL. SO NOW LET US TAKE ANY IRRATIONAL POINT A IN THE GIVEN INTERVAL FOR THE FUNCTION TO BE CONTINUOUS HERE WE SHALL HAVE LIMIT OF F[X] AS X TENDS TO A SHALL EXIST AND SHALL BE F[A].......................2 SINCE A IS IRRATIONAL , WE GET F[A]=0.......AS GIVEN NOW LET US CHECK THE LIMIT IS 0 OR NOT IT WILL BE THE LIMIT IF WE CAN MAKE |F[X]-0]|<E >0.................WHERE E IS A SPECIFIED SMALL NUMBER BY TAKING X AS CLOSE TO A AS DESIRED BUT FROM THE STATEMENT MADE AT THE BEGINING , THERE ARE INFINITE RATIONAL POINTS AROUND EVERY IRRATIONAL POINT. SO F[X] AT THAT POINT = P/Q = 1/Q AS PER HYPOTHESIS. SO WE GET ||[1/Q]-0|<E............POSSIBLE SINCE A= P/Q IS A VARIABLE NUMBER NEAR ZERO ...THAT IS Q COULD BE VERY LARGE WITH P=1 AND SO WE CAN DO IT BY TAKING X CLOSER TO A ....SAY BY MAKING Q=R>1/E, WHERE R IS AN INTEGER SO F[X] FAS LIMIT EQUAL TO ZERO AS X TENDS TO A AN IRRATIONAL NUMBER WHICH IS ALSO EQUAL TO F[A] SO THE FUNCTION IS CONTINUOUS AT ANY IRRATIONAL NUMBER IN THE GIVEN INTERVAL.
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