EIV 4th Edition Chapter 6 Prob 3 page 101

[georgepa]

If G^2 < G - which of the following could be G ...

The answer is 0 < G < 1

Could someone give me an algebraic solution to this?

I am doing something wrong here:

G^2 < G
=> G ^2 - G < 0
=> G( G - 1) < 0
=> G < 0 , G-1 < 0
=> G < 0, G < 1
=> G < 1 is obviously wrong


The other way makes more sense

G^2 < G
=> 0 < G - G^2
=> 0 < G(1-G)
=> 0 < G , 0 < 1-G
=> 0 < G , -1 < -G
=> 0 < G, 1 > G
=> 0 < G, G < 1
=> 0 < G < 1

Thanks,
/george

[georgepa]

May have found the answer

G(G-1) < 0

This means

1) G < 0 and (G-1) > 0

G < 1 and G > 1 -> Invalid

or

2) G >0 and (G-1) < 0

G > 0 and G < 1

0 < G and G < 1

0 < G < 1

for quadratic inequalities you need to test all cases

E.g. if

(x-a)(x-b) < 0

Then we know that either one of the terms is -ve so the two cases are

(x-a) <0 and (x-b) > 0

or

(x-a) >0 and (x-b) < 0

Similarly if

(x-a)(x-b) > 0

We know that both terms will have the same sign

So:

(x-a) > 0 and (x-b) > 0

or

(x -a) < 0 and (x-b ) <0

[Ben Ku]

Your solution is basically correct.

Let's look at it this way:
G(G-1) < 0

What this means is that G and G-1 have opposite signs.

Well, this inequality might be hard, but we do know the answer to the equation:
G(G - 1) = 0
G= 0 or 1

So 0 and 1 are boundary points that divide up our numbers into three regions:
<-----------0---------1---------->

Our three regions are:
(1) G < 0
(2) 0 < G < 1
(3) G > 1

If we test region (1),
G is negative
G - 1 is negative
so G(G-1) is positive.

If we test region (2),
G is positive
G - 1 is negative
so G(G-1) is negative.

If we test region (3),
G is positive
G - 1 is positive
so G(G-1) is positive.

Here, only the region 0 < G < 1 has solutions to our inequality. Hope that helps.