- #1
paridiso
- 17
- 0
proof question from "How to Prove It"
3.2.8 Suppose a & b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1
I understand intuitively why this is true, but I can't figure out how to prove it. According to the hints at the back of the book it says to prove a < 0, then use to prove a < -1.
When I go through the inequalities I come up with this:
a < 1/b
ab < 1
1/a < b
1 < ab
I know that when you multiply both sides of inequality you have to switch the signs. But if both a and b are negative the signs both switch so I don't really understand how this can be true.
3.2.8 Suppose a & b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1
I understand intuitively why this is true, but I can't figure out how to prove it. According to the hints at the back of the book it says to prove a < 0, then use to prove a < -1.
When I go through the inequalities I come up with this:
a < 1/b
ab < 1
1/a < b
1 < ab
I know that when you multiply both sides of inequality you have to switch the signs. But if both a and b are negative the signs both switch so I don't really understand how this can be true.