EIV Chap 6 Qn 3

[s1800028]

Hi,

Is there another way to solve this problem?

What I did was this:
G^2 < G
G(G-1) < 0
G < 0 or G < 1
Therefore G<0.

[Ben Ku]

Hi,

I don't think your approach to this question is quite right.

What I did was this:
G^2 < G
G(G-1) < 0
G < 0 or G < 1
Therefore G<0.

"G(G-1) < 0" does not necessarily lead to "G < 0 or G < 1". In fact, "G < 0 or G < 1" actually means G < 1 (since OR means you must combine both inequalities).

G(G-1) < 0 means that G and (G-1) have opposite signs The only region where this is true is between 0 and 1.

Another way to look at this problem is:
If G^2 < G, the only interval where it's true that the square is smaller than a number is between 0 and 1. (See Number Properties Strategy Guide page 65)

Hope that helps!

[pandoracarey]

I understand why this question works out the way that it does. But, in the answer key it says "If Gsquared < G, then G must be positive" i dont get that because G can also be a negative #. Do we know G is positive because we also know that "Gsquared < G"? I also understand that Gsquared will never be a negative # but my understanding is that G can be either neg or pos.

Thanks!

[messi10]

Hi pandoracarey,

When a square of a number is less than the number then we know that that number is between 0 and 1. If a number is between 0 and 1 then it automatically means that it will always be positive.

Regards

Sunil

[jnelson0612]

Great Sunil, exactly.