General Inequality question

[Guest]

Have a question about inequalities

if 1/(k-1) > 0 can I inverse this and say (k-1) > 0?

is that correct to do?

thanks

[RA]

For 1/(k-1) to be greater than 0, (k-1) has to be Positive. so the operation you performed appears valid to me.

[Guest]

There is nothing called as "inverse" in inequalities. You multiply or divide always..... and make sure to maintain sanity of the equation you need to mul/div by a positive number.
What you just did was multiply both sides by (k-1)^2. Since the square of any number is always positive the operation is valid.

[Guest]

think simple..

1/(k-1) > 0 means a positive number is divided by (k--1) and it still is positive SO...(k-1) is positive too

[esledge]

Good discussion.

Seeing zero on one side of an inequality is a strong message that you can approach it as a "sign" number property question, as suggested above:

think simple..

1/(k-1) > 0 means a positive number is divided by (k--1) and it still is positive SO...(k-1) is positive too

In contrast, if a non-zero integer is on one side of an inequality, you should/must take a more algebraic approach. An example:
1/(k-1) > -11
11+1/(k-1) > 0
11(k-1)^2+(k-1) > 0 (multiplying both sides by (k-1)^2, no sign flipping required, as this is definitely positive.)
11(k^2-2k+1)+(k-1) > 0
(11k^2-22k+11)+(k-1) > 0
11k^2-21k+10 > 0 (it's not likely the GMAT will make you factor something like this, but it's just what I made up.)
(11k-10)(k-1) > 0
From here, you can approach it as a "sign" number properties question. FYI, the solution is k<10/11 or k>1.