# why does (g^2 – 1)/(g – 1) = g + 1 ?

humy
Science 23 Sep '17 08:43
1. 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?
2. 23 Sep '17 09:041 edit
Originally posted by @humy
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?
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.
3. 23 Sep '17 09:283 edits
Originally posted by @fabianfnas
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.
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.
4. 23 Sep '17 11:37
Originally posted by @humy
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.
5. 23 Sep '17 11:52
Originally posted by @humy
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.
No one "solves" math questions; they merely discover what was already there.
6. 23 Sep '17 13:11
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).
7. 23 Sep '17 13:44
Originally posted by @kazetnagorra
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).
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.
8. 23 Sep '17 13:54
Originally posted by @humy
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?
Let me guess, you thought it should be g.
9. 23 Sep '17 15:211 edit
Originally posted by @humy
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.
Factoring and reducing is easier.

But this leads to something most people miss. They view the world only through their eyes. The parts of math they think important becomes math. So those who are good at what we value are deemed good at math while others who are good with other parts are deemed inferior.

I'd say the ones least likely to fall prey to this point of view are people who teach people not topics. If your goal is to get someone to understand based on how the individual sees things, you teach people. If you try to get people to understand how you do it, you teach a topic.

For math, University teachers generally teach topics. Many high school teachers teach topics. A few of us teach people. Many who claim to teach people just water things down, so it is possible to be in error in that group too.
10. wolfgang59
Infidel
23 Sep '17 16:51
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.
Of course they are equal.
g is not a variable, it is a constant not equal to 1.
Your smart arse thinking is bs
11. 23 Sep '17 18:38
Originally posted by @wolfgang59
Of course they are equal.
g is not a variable, it is a constant not equal to 1.
Your smart arse thinking is bs
Why would g be a constant not equal to one for g+1?
12. 23 Sep '17 19:291 edit
Why would g be a constant not equal to one for g+1?
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?
13. 23 Sep '17 19:39
Originally posted by @humy
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?
It is not a paradox. Strictly speaking they are not equal.

One is a linear equation which is defined for all real numbers, the other is a rational equation which is defined for all real numbers except 1.

They just happen to have the exact same answers everywhere except at x equals one.

The proper understanding is that they are equivalent at all real numbers except 1. Equivalent and equal are not the same thing, but what kids are taught leads people to these faulty assumptions.
14. 23 Sep '17 21:43
Originally posted by @humy
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?

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.
15. 23 Sep '17 21:51
Originally posted by @kazetnagorra