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Re: AB positive or negitive [#permalink]
31 Jan 2011, 01:17
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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: AB positive or negitive [#permalink]
02 Feb 2011, 05:02
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.
Re: AB positive or negitive [#permalink]
02 Feb 2011, 05:33
Expert's post
2
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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.
Re: If a b > a + b, where a and b are integers, which of the [#permalink]
15 Oct 2014, 15:22
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Re: If a b > a + b, where a and b are integers, which of the [#permalink]
24 Aug 2015, 08:55
Expert's post
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 ?
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]
24 Aug 2015, 08:56
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. _________________
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