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?

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23 Sep '17 09:041 edit

Trying:*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?**

(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.- Joined
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23 Sep '17 09:283 edits

thanks for that. Now it's just a simple matter of expanding (g+1)(g-1) to show RHS LHS equivalence thus;*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.

(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.- Joined
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23 Sep '17 11:37

Glad I could help you. 🙂*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.- Joined
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Territories Unknown23 Sep '17 11:52

No one "solves" math questions; they merely discover what was already there.*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.- Joined
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Germany- Joined
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23 Sep '17 13:44

That was what I was going to point out. They are not equal.*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).**

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.- Joined
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23 Sep '17 13:54

Let me guess, you thought it should be g.*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?**- Joined
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23 Sep '17 15:211 edit

Factoring and reducing is easier.*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.

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.- Joined
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Dunedin23 Sep '17 16:51

Of course they are equal.*Originally posted by @eladar***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.

g is not a variable, it is a constant not equal to 1.

Your smart arse thinking is bs- Joined
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23 Sep '17 18:38

Why would g be a constant not equal to one for g+1?*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- Joined
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23 Sep '17 19:291 edit

g can equal 1 for g+1 but not for (g^2 – 1)/(g – 1) .*Originally posted by @eladar***Why would g be a constant not equal to one for g+1?**

And yet g+1 = (g^2 – 1)/(g – 1).

ANYONE;

Is there a paradox somewhere there?- Joined
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23 Sep '17 19:39

It is not a paradox. Strictly speaking they are not equal.*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?

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.- Joined
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Germany23 Sep '17 21:43

As Eladar correctly points out, there is no paradox.*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.- Joined
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23 Sep '17 21:51

I think more people would get it if it was stated like this...*Originally posted by @kazetnagorra***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.

As long as g does not equal one, g+1 results in the same answers as (g^2-1)/(g-1).