Show a). 1>0 b). if 0<a<b, then 0<1/b<1/a c).If 0<a<b, then a^n < b^n for any po
ID: 2968302 • Letter: S
Question
Show
a). 1>0
b). if 0<a<b, then 0<1/b<1/a
c).If 0<a<b, then a^n < b^n for any positive integer n
d).If a and b are positive numbers, and a^n < b^n for some positive integer n, show a<b
e).Show for any real numbers a and b that lal <=lbl iff a^2 <= b^2
f). Let x and y be positive real numbers with x<y and let s be a positive rational number, then x^s < y^s
Explanation / Answer
a) If x>0 , x^2>0 . IF x<0, then -x>0 hence (-x)^2>0. But x^2= (-x)^2 by (d) Since 1=1^2, 1>0
b) 0<a<b
Divide both sides by ab to get 0/ab < a/ab < b/ab or 0<1/b<1/a
c) 0<a<b
So a<b
Taking nth power of both sides
a^n<b^n (As n is positive )
d) a^n < b^n
Taking 1/n th power of both sides, a^n*1/n < b^n*1/n
=> a < b
e) sqrt (a^2) = lal and sqrt (b^2) = ( lbl)
Now a^2 <= b^2 So taking sqrt of both sides we get lal <= lbl
f) x<y
Taking sth power of both sides x^s<y^s [As s is positive rational]