Answer:
a. true b. true
Step-by-step explanation:
(a) If r and s are rational numbers, then (r+s)/2 is rational.
true
rational numbers can be expresed as fractions
let be r=a/b and s=c/d being a,b,c,d integer numbers
[tex]\frac{r+s}{2} =\frac{\frac{a}{b}+\frac{c}{d} }{2} =\frac{\frac{da+bc}{bd} }{2} =\frac{da+bc}{2bd}[/tex]
d.a=e is an integer number because it's the product of two integers
b.c=f is an integer number because it's the product of two integers
e+f=g is an integer number because it's the sum of two integers
b.d=h is an integer number because it's the product of two integers
2.h=i is an integer number because it's the product of two integers
g/i=j is an integer number because it's the quotient of two integers
then
[tex]\frac{r+s}{2} =\frac{\frac{a}{b}+\frac{c}{d} }{2} =\frac{\frac{da+bc}{bd} }{2} =\frac{da+bc}{2bd}=\frac{e+f}{2h} =\frac{g}{i} =j[/tex]
(b) For all real numbers a and b, if a < b then a < (a+b)/2 < b
true
[tex]a < (a+b)/2 < b[/tex]
[tex]2a < (a+b) < 2b[/tex]
lets analyze 2a < (a+b)
[tex]2a < (a+b) \\2a-a < (a+b)-a\\a < b[/tex]
then 2a < (a+b) is true
lets analyze (a+b) < 2b
[tex](a+b) < 2b\\(a+b)-b < 2b-b\\a< b[/tex]
then (a+b) < 2b is true
