- #1
MidgetDwarf
- 1,480
- 616
Prove that if 0<a<b, then [itex]a < \sqrt{ab} < \frac {a+b} {2} < b [/itex]
Please excuse if format is messy, this is my first time writing in Latex.
Suppose 0<a<b then 0<a+O =a (1). Since a<b we can can rewrite this using (1) which is a+0<b (2).
Adding a to both sides of (2) (closed under addition) we get 2a< a+b (3). Using the multiplicative inverse of 2, (3) now becomes ## a < \frac {a + b} {2} ## (4). Using the fact that a<b, ## a < \frac {a + b} {2} ##
< ## a < \frac {a + b} {2} ## < ## a < \frac {b + b} {2} ## =b. Therefore a<b .
Here is were I am confused. I am missing the step were ## a < \sqrt {ab} ##. Can I say that a*b<a*a=a^2 ?
Then take the square root? If this is true how would I connect it with ## a < \frac {a + b} {2} ## ?
Strong criticism welcomed.
Please excuse if format is messy, this is my first time writing in Latex.
Suppose 0<a<b then 0<a+O =a (1). Since a<b we can can rewrite this using (1) which is a+0<b (2).
Adding a to both sides of (2) (closed under addition) we get 2a< a+b (3). Using the multiplicative inverse of 2, (3) now becomes ## a < \frac {a + b} {2} ## (4). Using the fact that a<b, ## a < \frac {a + b} {2} ##
< ## a < \frac {a + b} {2} ## < ## a < \frac {b + b} {2} ## =b. Therefore a<b .
Here is were I am confused. I am missing the step were ## a < \sqrt {ab} ##. Can I say that a*b<a*a=a^2 ?
Then take the square root? If this is true how would I connect it with ## a < \frac {a + b} {2} ## ?
Strong criticism welcomed.