23 Sep '17 08:43>
not sure if I am being stupid here but I so far failed to see why. Can someone show me the intermediate algebraic expressions that show how you can go from (g^2 – 1)/(g – 1) to g + 1?
Originally posted by @humyTrying:
not sure if I am being stupid here but I so far failed to see why. Can someone show me the intermediate algebraic expressions that show how you can go from (g^2 – 1)/(g – 1) to g + 1?
Originally posted by @fabianfnasthanks for that. Now it's just a simple matter of expanding (g+1)(g-1) to show RHS LHS equivalence thus;
Trying:
(g^2 – 1)/(g – 1) = (g+1)
Multiply right and left with (g-1), knowing that (g+1)(g-1)= (g^2-1) and you have your identity.
I suppose g doesn't equal 1.
Originally posted by @humyGlad I could help you. 🙂
thanks for that. Now it's just a simple matter of expanding (g+1)(g-1) to show RHS LHS equivalence thus;
(g+1)(g-1) = g^2 - g + g - 1 = g^2 - 1
Its strange how I often solve the most complex maths problems but fail with the simplest.
Originally posted by @humyNo one "solves" math questions; they merely discover what was already there.
thanks for that. Now it's just a simple matter of expanding (g+1)(g-1) to show RHS LHS equivalence thus;
(g+1)(g-1) = g^2 - g + g - 1 = g^2 - 1
Its strange how I often solve the most complex maths problems but fail with the simplest.
Originally posted by @kazetnagorraThat was what I was going to point out. They are not equal.
Generally one can write x² - a² = (x-a)(x+a). Hence g² - 1 = (g-1)(g+1) and the equality follows (provided g does not equal 1).
Originally posted by @humyLet me guess, you thought it should be g.
not sure if I am being stupid here but I so far failed to see why. Can someone show me the intermediate algebraic expressions that show how you can go from (g^2 – 1)/(g – 1) to g + 1?
Originally posted by @humyFactoring and reducing is easier.
thanks for that. Now it's just a simple matter of expanding (g+1)(g-1) to show RHS LHS equivalence thus;
(g+1)(g-1) = g^2 - g + g - 1 = g^2 - 1
Its strange how I often solve the most complex maths problems but fail with the simplest.
Originally posted by @eladarOf course they are equal.
That was what I was going to point out. They are not equal.
They are different at one point, therefore are not equal.
Exclude that one point then they are equal.
The same kind of screwed up thinking makes people think the square root of x squared equals x.
Originally posted by @wolfgang59Why would g be a constant not equal to one for g+1?
Of course they are equal.
g is not a variable, it is a constant not equal to 1.
Your smart arse thinking is bs
Originally posted by @eladarg can equal 1 for g+1 but not for (g^2 – 1)/(g – 1) .
Why would g be a constant not equal to one for g+1?
Originally posted by @humyIt is not a paradox. Strictly speaking they are not equal.
g can equal 1 for g+1 but not for (g^2 – 1)/(g – 1) .
And yet g+1 = (g^2 – 1)/(g – 1).
ANYONE;
Is there a paradox somewhere there?
Originally posted by @humyAs Eladar correctly points out, there is no paradox.
g can equal 1 for g+1 but not for (g^2 – 1)/(g – 1) .
And yet g+1 = (g^2 – 1)/(g – 1).
ANYONE;
Is there a paradox somewhere there?
Originally posted by @kazetnagorraI think more people would get it if it was stated like this...
As Eladar correctly points out, there is no paradox.
We can write the equation as:
(g+1)*(g-1)/(g-1) = g+1
It is easy to see that the equality holds if (g-1)/(g-1) = 1. Dividing something by itself will always give 1 except when you are dividing 0 by itself, which is undefined. Hence the equality holds only when g - 1 is not zero, i.e. when g is not 1.