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Unless you know whether a is positive or not you cannot determine how the inequality will look like after you multiply it by a. You cannot still prove III to be true.

Lets consider the two cases here: Suppose a is positive (3, say) 3-b>3+b Multiply the inequality by 3 9-3b>9+3b 6b<0 b<0

hence, ab<0

Now, suppose a=-3 plug in the values:

-3-b>-3+b Multiply by a(-3) 9+3b<9-3b 6b<0 b<0

hence, ab > 0

Thus the sign of ab will vary depending on the sign of a. Hope this clears up the confusion

Re: If a b > a + b, where a and b are integers, which of the [#permalink]

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18 Jul 2017, 11:10

1

This post received KUDOS

I definitely found this confusing, but let me try and explain how I went through this.

a - b > a + b --> subtract a from both sides. - b > b --> b is clearly negative. 0 > 2b 0 > b --> yup, negative; (II) is correct.

What else?

Well can 'a' be positive? Yes, a positive number (-) a negative number is always greater than the sum of two negative numbers. Can a be negative? Ok, well what if a = -1...

In this case, the answer to both (I) and (III) is 'NO' as a could be positive or negative.

More conceptually however, it might be more edifying to convince oneself that a negative number minus another negative number (so the negative cancel out and it becomes addition) is greater (less negative) and a negative number plus another negative number (which just goes more negative).

Q: If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

Through inequality; I just derived that b is -ve a – b > a + b -b > b --> subtracting 'a' from both sides -2b > 0 --> subtracting 'b' from both sides -b > 0 --> dividing both sides by 2 b < 0 --> multiplying -ve sign on both sides

The rest two; I ruled out using numbers

Let b, as we know is -ve, to be equal to -1

Case I a=+ve, say 100 a - b = 100 – (-1) = 101 a + b = 100 + (-1) = 99

a-b > a+b

Case II a=-ve, say -100 a - b = -100 – (-1) = -100 + 1 = -99 a + b = -100 + (-1) = -100 - 1 = -101

So, a - b > a + b

Thus, a - b > a + b is true for both +ve and -ve 'a'

We just proved that a-b>a+b is true for both +ve and -ve values of a.

We can't conclusively say that a < 0. Statement I is ruled out.

for a=+ve; ab = +ve * -ve = -ve and for a=+ve; ab = -ve * -ve = +ve

So, we can't conclusively say ab < 0. Statement III is ruled out.

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A. I only B. II only C. I and II only D. I and III only E. II and III only

Merging topics.

Please refer to the discussion above.

can we not sqaure both sides of a – b > a + b ?

1. No. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). Check for more here: inequalities-tips-and-hints-175001.html

Re: If a b > a + b, where a and b are integers, which of the [#permalink]

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24 Aug 2015, 09:56

anupamadw wrote:

Bunuel wrote:

shasadou wrote:

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A. I only B. II only C. I and II only D. I and III only E. II and III only

Merging topics.

Please refer to the discussion above.

can we not sqaure both sides of a – b > a + b ?

You are only complicating the simple expression given by squaring it.

Also, as a general rule of inequalities, squaring variables in the inequalities is not a good idea until you know for sure the sign of a-b as multiplication by a negative number reverses the inequality.

If a b > a + b, where a and b are integers, which of the [#permalink]

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19 Feb 2016, 22:40

Bunuel wrote:

girishkakkar wrote:

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

Answer: B.

Hi Bunuel,

I cannot understand, how could we cancel 'a' here when we do not know about its sign? Please explain this.

If a – b > a + b, where a and b are integers, which of the following must be true?

I. a < 0 II. b < 0 III. ab < 0

A I only B II only C I and II D I and III E II and III

You don't need number plugging at all.

Given: \(a-b>a+b\) --> \(a\) cancels out (which means that from given info we can say nothing about it: it can be positive, negative or zero, so it can be ANY number) --> \(2b<0\) --> \(b<0\).

As we know nothing about \(a\) and know that \(b<0\), so only II must be true.

Answer: B.

Hi Bunuel,

I cannot understand, how could we cancel 'a' here when we do not know about its sign? Please explain this.

Thanks.

There is a difference between reducing (dividing/multiplying) an inequality by a variable and subtracting/adding a variable to both parts of an inequality. The first one we cannot do unless we know the sign of the variable but subtracting/adding is always applicable.