Just Compare

$\frac{A}{A-B} - \frac{B}{A-B} = \frac{A-B}{A-B} = 1,$ So, the first value always equals the second value plus one, thus the first value is greater.

$\frac{A}{A-B} - \frac{B}{A-B} = \frac{A-B}{A-B} = 1$ thus $\frac{A}{A-B} > \frac{B}{A-B}$

If $A > B$ , the denominator is positive, and is the same denominator for both. As $A > B$ we conclude that the first one is greater.

If $A < B$ , the denominator is negative, and the same for both.

Case 1 If $A$ is positive, $B$ is forced to be positive and hence, the expressions are negative. Since $|A| > |B|$ , this case claims the first expression as the greater.

Case 2 If $A$ is $0$ , the first expression is $0$ , and as $B$ is forced to be positive, the second expression is negative. $0$ > than any negative number.

Case 3 If $A$ is negative, there are three sub-cases:

Sub-Case a: If $B$ is positive, the first expression is positive, and the second is negative.

Sub-Case b: If $B$ is $0$ , the first expression is positive and the second is $0$ .

Sub-Case c: If $B$ is negative, both expressions turn to be positive. Remember that the denominator is always negative because $A < B$ . As $|A| > |B|$ , this claims again the first expression as the greater.

We've evaluated and proved that the first expression is always greater than the other. Sorry if there's some mistake.

Just subtract First -Second = ( A-B)/(A-B)=1 So First must be greater.

Suppose, first, that $A > B$ . Then $\frac{1}{A-B}$ is a positive number. We write

$A \times \frac{1}{A-B} > B \times \frac{1}{A-B}$

Now suppose $B > A$ . Then $\frac{1}{A-B}$ is a negative number and we can write

$A \times \frac{1}{A-B} > B \times \frac{1}{A-B}$

So, either way, $A \times \frac{1}{A-B} > B \times \frac{1}{A-B}$ .

$\frac{A}{A-B}-\frac{B}{A-B}=1$

Given that the first minus the second is equal to 1 then the first is greater $\frac{A}{A-B} - \frac{B}{A-B} = \frac{A-B}{A-B}=1$

First note that if $A \neq B \quad then \quad {{A - B} \over {A - B}} = 1$

If A > B , then:

${{A - B} \over { A - B}} = 1 > 0 \quad \to \quad {A \over {A - B}} - {B \over {A - B}} > 0 \quad \to \quad {A \over {A - B}} > {B \over {A - B}}$

first - second=1>0 ergo first >second

subtract the second from the first , you get =1 i.e. (A/(A-B))-(B/(A-B) ) = ( (A-B)/(A-B))= 1 [ given , a <> b, and therefore a-b <>0]

therefore expression 1 is > expression 2

Let A>B

Let A = 5 ; B = 4

Obviously 5 > 4

Now let A<B

let A = 4 ; B = 5

Again -4 > -5

In both cases The FIRST one is correct.........

If $A>B$ , then $A-B>0$ , which implies $\frac{A}{A-B}>\frac{B}{A-B}$ .

If $A < B$ , then $A-B<0$ , which implies $\frac{A}{A-B}>\frac{B}{A-B}$ . (when divide by negative number, it'll switch the sign.)

Therefore, $\frac{A}{A-B}>\frac{B}{A-B}$ , for any $A,B; A\neq B$ .

For all real numbers $A$ and $B$ , it's clear that $A^{2}+B^{2}\geq2AB$ . Since $A$ is not equal to B, then we get $A^{2}+B^{2}>2AB$ . Then $A^{2}+B^{2}>2AB$ $A^{2}-AB>AB-B^{2}$ $A(A-B)>B(A-B)$ Divide both sides with $(A-B)^{2}$ , then we get $\frac{A}{A-B}>\frac{B}{A-B}$ .

Here, let's take two possibilities. A>B and B>A

First scenario :

If A>B, hence, A/(A-B) > B/(A-B) [since A-B is positive]

If B>A, Then -B<-A. Which means -B/(B-A)<-A/(B-A) [since B-A is positive]

We see that in both cases, A/(A-B) is greater. This is how i understand the problem. There can be many other ways! Thanks!

For those who still not understand. Look! A is always greater! Lets assume A-B = x

1. For example, A is greater than B(i said greater, i dont care if they are negative or positive number, the important thing is A greater than B). Then x will be positive. A/x is then greater than B/x or A/x is LESS NEGATIVE than B/x

2. Now lets assume B is greater than A. Then x will be negative. B/x will be MORE NEGATIVE than A/x or B/x will be smaller than A/x.

No matter what number you put in, you will always get A/x bigger than B/x.

Notes : remember that we have to find which one is bigger, either A/x or B/x, not A or B!

I actually solved the problem a simple way.

Let A=2 B=1

then plug those values in the First and Second stuff, and First was greater than Second, so yah.

$\dfrac A{ A - B } =\dfrac {A - B + B}{ A - B } =\dfrac {A - B }{ A - B } +\dfrac B{ A - B } =1 + \dfrac B{ A - B }\\ 1st.$

I thought the tricky part would arise when I made A a negative number. But I tried a simple example and it turned out the same as if both numbers were positive. If A is -5 and B is 2, then the first fraction becomes -5/-7, which is 5/7. The second fraction becomes 2/-7, which is a neg number, -2/7. So as usual, the former is higher than the latter.

We just don't give a f*ck if A and B are positive or negative because First - Second always = (A-B)/(A-B) = 1. --> First = Second + 1 --> First > Second.

I think you should specify that a low negative number is considered greater than a high negative number. e.g. -2 > -4

First - Second = A/(A-B) - B/(A-B) = (A-B)/(A-B) = 1. So first one is always greater by 1 than second.

$\frac { A }{ A-B } >\frac { B }{ A-B } \\ \Rightarrow \frac { A }{ A-B } -\frac { B }{ A-B } >0\\ \Rightarrow \frac { A-B }{ A-B } >0\\ \forall A,B\in { Z }.\frac { A-B }{ A-B } =1\\ \Rightarrow 1>0\\ \therefore \frac { A }{ A-B } >\frac { B }{ A-B } \\$

1st - 2nd = 1

=> 1st = 1 + 2nd

=> Therefore; 1st > 2nd

if A>B then the first one is the greater , they have the same denominator but the Numerator of the first is greater so the first one is greater..... if A<B then they are both have negative sign and the first one is greater

if you subtract the second from the first, you get +1, which is clearly greater than zero. Thus, the first one is greater than the second one

It will always be the first even if B>A, sure the second |quantity| would be bigger but there will be a negative sign which will lower its value

ahmm I Do Not solve it but I answerd it right hahahaha I just thought that A is greater than B because A is not equal to B ?? Im right or wrong??

Put any value of B we get negative answer but put any value of A we get possitve answer so the FIRST is greater

Lets take (A-B) = x So A= x+B. B= A-x We have either A/x > B/x or the oppisite So if B/x is bigger we have (x+b)/x <B/x

X+B-B/x <0 we have x/x <0 which means 1 is < 0 If A/x is bigger we have (X+B)/x >B/x then we have

X+B-B/x >0 X/X>0 which means 1>0 therefore A/x have to be bigger or A/A-B

The first. If B is larger than A, then A-B is negative so both expressions are negative, so the number with the smaller modulus is larger, this number is the number with the smaller numerator (A) so the first expression is larger. If B is smaller than A then A-B is positive so the expressions are both positive, so the number with the larger modulus is larger, this is the number with the larger numerator (A) so the first expression is largest. Hence in all possible cases the first expression is larger.

A/(A-B) - B/(A-B) = 1 > 0 => A/(A-B)>B/(A-B)

Since

A/(A - B) - B/(A - B) = 1

Then

A/(A - B) > B/(A - B)

A/A-B < A

by above A<A(A-B)

1<A-B

1+B<A means

A>B

dividing b/s by A-B

A/(A-B)>B/(A-B)

Take any example e.g; A=5 and B = 4 and compare ,then Reverse the values, and compare both. One of the ways but not the best !!!

\frac{A}{A-B} - \frac{B}{A-B}= 1 So, the first value is greater

if a = n and b = n+1 then by substitution you get n/-1 for the first and (n+1)/-1 for the second. so the second is more negative therefore smaller!

A/A-B is a negative number and hence the smaller negative number will be greater than the higher negative number.

By sheer logic:

If A is bigger, The first be a smaller negative number than the second, so the first is bigger

If B is bigger, The first will be a larger positive value than the second.

just subtract the two fractions. first minus second gives one while the second minus first gives -1. second - first<0 implies second < first

First - Seccond = 1 >0 => First > Second

saying that A is not equal to B may refer to " A < B " so .. A-B is a negative value .. and because $A<B$
then less negative is greater .. so .. the first is greater ..

An MCQ can always be answered using examples take (A,B)=(1,2) and (A,B)=(2,3)..and we have the answer.

Case 1 :If A < B. Therefore, (A - B) < 0 {negative term} So, A/(A - B) > B/(A - B) . {As, 2 < 3, -2 > -3}.

Case 2: If A > B. Therefore, (A - B) > 0. So, A/(A - B) > B/(A - B)

Case 3: A cannot be equal to zero. {Given}

CASE 1:If A>B , then A/A-B > B/A-B , since both have same denominator but the numerator A is greater.

CASE 2: If B>A , then A-B is negtaive. So A/A-B > B/A-B, since though B/A-B will be greater in magnitude but the negative sign will make it smaller.

Let A>B Then it is obvious that A/A-B>B/A-B Let B>A Then -A/A-B<-B/A-B =>A>B Therefore First value will always be greater than the second one

if A>B,

A/A-B >0 and A/A-B >B/A-B

If B>A,

A/A-B <0 and again A/A-B>B/A-B (negative fractions)

Hence the first answer!

(A-B) is negative and A>B. So, A/(A-B) >B/(A-B) Like, let A=5 and B=6 so that A-B= -1 and the A/(A-B)=-5 and B/(A-B)= -6 In this case -5>-6

lets assume , A=5 B=6 , -5 > -6 ,[first condition is B>A] again A=6 B=5 , still 6>5 [ second condition being A>B], so first one is always greater.

when A>B FIRST>SECOND; WHEN B>A FIRST WILL BE LESS -VE NO..................SO FIRST>SECOND