Is the argument sound? Yes, the argument is sound.

[Keen94]

Homework Statement

Suppose that a and b are nonzero real numbers. Prove that if a< 1/a < b < 1/b then a<-1.

Homework Equations

Givens: a and b are nonzero real numbers, a< 1/a < b < 1/b, and a≥-1.
Goal: Arrive at a contradiction.

The Attempt at a Solution

Scratch work: First establish whether a<0 or a>0.
Case i.) a>0 and Case ii.) a<0
Case i). a>0. a( a<1/a) ⇒ a²<1 and a(a≥-1) ⇒ a²≥-a. However a²≥0 and -a<0. Therefore a>0 contradicts the fact that a is at leas greater than or equal to -1.
Is the argument sound? Thanks for the feedback!

 

[Mark44]

Homework Statement

Suppose that a and b are nonzero real numbers. Prove that if a< 1/a < b < 1/b then a<-1.

Homework Equations

Givens: a and b are nonzero real numbers, a< 1/a < b < 1/b, and a≥-1.
Goal: Arrive at a contradiction.

The Attempt at a Solution

Scratch work: First establish whether a<0 or a>0.
Case i.) a>0 and Case ii.) a<0
Case i). a>0. a( a<1/a) ⇒ a²<1 and a(a≥-1) ⇒ a²≥-a. However a²≥0 and -a<0. Therefore a>0 contradicts the fact that a is at leas greater than or equal to -1.
Is the argument sound? Thanks for the feedback!

Your argument doesn't do anything with b, which could also be either negative or positive.

Your hypothesis is (in part) that a < 1/a and that b < 1/b. What do these inequalities say about the numbers a and 1/a and b and 1/b?

Your other hypothesis is that a ≥ -1.

 

[vela]

Homework Statement

Suppose that a and b are nonzero real numbers. Prove that if a< 1/a < b < 1/b then a<-1.

Homework Equations

Givens: a and b are nonzero real numbers, a< 1/a < b < 1/b, and a≥-1.
Goal: Arrive at a contradiction.

The Attempt at a Solution

Scratch work: First establish whether a<0 or a>0.
Case i.) a>0 and Case ii.) a<0
Case i). a>0. a( a<1/a) ⇒ a²<1 and a(a≥-1) ⇒ a²≥-a. However a²≥0 and -a<0. Therefore a>0 contradicts the fact that a is at leas greater than or equal to -1.
Is the argument sound? Thanks for the feedback!

No. Take $a=1/2$ for example. Clearly, $a>0$ holds so $a \ge -1$ does as well. Moreover, you have $a^2 = 1/4 \ge 0$ and $-a = -1/2 < 0$. There's no apparent contradiction.

 

[HallsofIvy]

Homework Statement

Suppose that a and b are nonzero real numbers. Prove that if a< 1/a < b < 1/b then a<-1.

Homework Equations

Givens: a and b are nonzero real numbers, a< 1/a < b < 1/b, and a≥-1.
Goal: Arrive at a contradiction.

The Attempt at a Solution

Scratch work: First establish whether a<0 or a>0.
Case i.) a>0 and Case ii.) a<0
Case i). a>0. a( a<1/a) ⇒ a²<1 and a(a≥-1) ⇒ a²≥-a. However a²≥0 and -a<0. Therefore a>0 contradicts the fact that a is at leas greater than or equal to -1.
Is the argument sound? Thanks for the feedback!

I don't see any where in here where you establish the "fact" that "a is greater than or equal to -1" that you say is contradicted. Of course, in the case that a> 0 it certainly must be greater than -1, but that doesn't seem to be what you are referring to. You do show, correctly, that a²≥-a. But that's trivially true if x> 0. It certainly doesn't contradict "a is greater than or equal to -1"!

 

[Keen94]

I see I forgot to post the solution here. woops sorry guys.

If a,b≠0 and a<1/a<b<1/b then a<-1
We can see that
a<1/b ⇒ab<1 if b>0 or it implies that ab>1 if b<0
and
1/a<b ⇒ab<1 if a<0 or it implies that ab>1 if a>0
We can conclude that ab<1 when b>0 and a<0 otherwise the inequality a<b does not hold.
Since a<1/a
then a²>1 since a<0
therefore lal>1
⇒a>1 or a<-1 bur a<0 thus a<-1.