Thread: Can't figure out this identity problem

1. Can't figure out this identity problem

Hello, I'm studying for the GRE and i'm trying to figure out how (1/b-1/a)=(a-b/ab). If anyone can demonstrate the steps of how these two expressions are identical I would greatly appreciate it. Thanks.

2. Re: Can't figure out this identity problem

common denominator is $ab$ ...

$\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{a}{ab} - \dfrac{b}{ab} = \dfrac{a-b}{ab}$

3. Re: Can't figure out this identity problem

No, no one can explain "how these two expressions are identical" because they are not! What is true is that 1/b- 1/a= (a- b)/ab. To do that, get "common denominators" in the two fractions: (a/a)(1/b)- (b/b)(1/a)= a/ab- b/ab= (a- b)/ab.

4. Re: Can't figure out this identity problem

Originally Posted by rgragsda
Hello, I'm studying for the GRE and i'm trying to figure out how (1/b-1/a)=(a-b/ab).
Originally Posted by HallsofIvy
No, no one can explain "how these two expressions are identical" because they are not! What is true is that 1/b- 1/a= (a- b)/ab. To do that, get "common denominators" in the two fractions: (a/a)(1/b)- (b/b)(1/a)= a/ab- b/ab= (a- b)/ab.
Both of you need grouping symbols around your product denominators.

1/b - 1/a = (a - b)/(ab)

(a/a)(1/b) - (b/b)(1/a)= a/(ab) - b/(ab) = (a - b)/(ab)

5. Re: Can't figure out this identity problem

Originally Posted by greg1313
Both of you need grouping symbols around your product denominators.

1/b - 1/a = (a - b)/(ab)

(a/a)(1/b) - (b/b)(1/a)= a/(ab) - b/(ab) = (a - b)/(ab)
... a plug for Latex?