Solving an algebraic identity with ellipses

[opus]

Homework Statement

Prove the following relation. It is assumed that all values of x and y which occur are such that the denominators in the indicated fractions are not equal to 0.

$$\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+...+x+1$$

Homework Equations

The Attempt at a Solution

Please see attached document for my work and thought process. Will type it out if it isn’t clear to anyone.
I don’t know how to continue with step (4) as I don’t think I understand the nature of the expression with the ellipses etc and how I would go about distributing to cancel terms.

 

[Mark44]

Homework Statement

Prove the following relation. It is assumed that all values of x and y which occur are such that the denominators in the indicated fractions are not equal to 0.

$$\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+...+x+1$$

Homework Equations

The Attempt at a Solution

Please see attached document for my work and thought process. Will type it out if it isn’t clear to anyone.
I don’t know how to continue with step (4) as I don’t think I understand the nature of the expression with the ellipses etc and how I would go about distributing to cancel terms.

Your work might not be enough. In the step where you multiply x - 1 and $x^{n-1} + x^{n-2} \dots + x + 1$, you are tacitly assuming that the left and right sides are equal, but this is what you are supposed to show (i.e., prove).

One way to show that the left side equals the right side is to perform polynomial long division, but this involves a bit of hand-waving. If you know how to do a proof by induction, that would be the ideal method to use.

Regarding the work in the image, you have
$x - 1(x^{n - 1} + x^{n-2} + \dots + x + 1)$, which is incorrect. It should be $(x - 1)(x^{n - 1} + x^{n-2} + \dots + x + 1)$.
In your step 4, just multiply using the distributive property -- each term in the binomial has to multiply each term in the longer polynomial. Again, though, your instructor might not consider this a valid proof.

 

[Ray Vickson]

Homework Statement

Prove the following relation. It is assumed that all values of x and y which occur are such that the denominators in the indicated fractions are not equal to 0.

$$\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+...+x+1$$

Homework Equations

The Attempt at a Solution

Please see attached document for my work and thought process. Will type it out if it isn’t clear to anyone.
I don’t know how to continue with step (4) as I don’t think I understand the nature of the expression with the ellipses etc and how I would go about distributing to cancel terms.

Ellipses are used to say "continue with that pattern", and are useful when a pattern is apparent, as in this problem. Sometimes the only other way of writing some expressions without using ellipses would be to use something like summation notation instead, such as $\sum_{k=0}^n x^k$. We need some way to write such summations when the number of terms is not specified, and sometimes even when the number of terms is specified but is large. For example, would you rather write out all 101 terms of $\sum_{k=0}^{100} x^k$ rather than $1+x+x^2 + \cdots + x^{99} + x^{100}?$

 

[epenguin]

With ellipses?

 

[Mark44]

With ellipses?

An ellipse is a geometric figure. An ellipsis is the three dots - ... -.
The plural of both is ellipses.

 

[epenguin]

An ellipse is a geometric figure. An ellipsis is the three dots - ... -.
The plural of both is ellipses.

This merits an entry in "Today I Learned".

 

[Mark44]

The thread title really should have ended with "with an ellipsis." At first I thought the question had to do with the geometric figure.

 

[opus]

Your work might not be enough. In the step where you multiply x - 1 and $x^{n-1} + x^{n-2} \dots + x + 1$, you are tacitly assuming that the left and right sides are equal, but this is what you are supposed to show (i.e., prove).

One way to show that the left side equals the right side is to perform polynomial long division, but this involves a bit of hand-waving. If you know how to do a proof by induction, that would be the ideal method to use.

Regarding the work in the image, you have
$x - 1(x^{n - 1} + x^{n-2} + \dots + x + 1)$, which is incorrect. It should be $(x - 1)(x^{n - 1} + x^{n-2} + \dots + x + 1)$.
In your step 4, just multiply using the distributive property -- each term in the binomial has to multiply each term in the longer polynomial. Again, though, your instructor might not consider this a valid proof.

I'm very new to the whole proof process thing, as this is a basic high school math book. I thought I could just smack it a few times with legal algebraic manipulations and if the two sides end up being identical, then the given equation is a true statement? You're saying this is not an acceptable way to do a proof?

I didn't think about dividing the LHS via long division. The $x^n$ throws me for a loop there. I'm not sure how I'd go about divining them to end up with something on the RHS as I don't see the pattern in $x^{n-1} + x^{n-2} \dots + x + 1$
I look at this as going $x^{n-1}+x^{n-2}+x^{n-3}+x^{n-4}\dots+x+1$ How can I look at this as being something that I can get through division on the LHS?

And again, in terms of step 4 an distributing though, I don't see how I can do this with the missing terms. I know I'm not looking at it correctly as I think that the ellipses means that the starting pattern will go on forever.

 

[Mark44]

I'm very new to the whole proof process thing, as this is a basic high school math book. I thought I could just smack it a few times with legal algebraic manipulations and if the two sides end up being identical, then the given equation is a true statement? You're saying this is not an acceptable way to do a proof?

When you multiplied x - 1 and $x^{n -1} + x^{n - 2} + \dots + x + 1$, you were tacitly assuming that $\frac{x^n - 1}{x - 1} = x^{n -1} + x^{n - 2} + \dots + x + 1$. This assumption isn't automatically correct, as long as the legal algebraic manipulations (your terminology) you do are reversible.

Here is sort of an example. I say "sort of" because it's solving an equation rather than proving an equation is true.
$\sqrt{x - 1} = -2$
Square both sides:
$x - 1 = (-2)^2 = 4$
$x = 5$ Tada!
Except that $\sqrt 5 \neq -2$
The problem is that squaring both sides of an equation is not a reversible operation for the reason that $f(x) = x^2$ is not a one-to-one function. (For a given output value, there are generally two input values.

Your proof would be acceptable if the "legal algebraic manipulations" you applied were limited to adding (subtracting) the same number to (from) both sides, multiplying both sides by the same nonzero number, dividing both sides by the same nonzero value, and a few more operations that are one-to-one (like taking logs, raising both sides as a power of, say, e, and a few others).

I didn't think about dividing the LHS via long division. The $x^n$ throws me for a loop there. I'm not sure how I'd go about divining them to end up with something on the RHS as I don't see the pattern in $x^{n-1} + x^{n-2} \dots + x + 1$
I look at this as going $x^{n-1}+x^{n-2}+x^{n-3}+x^{n-4}\dots+x+1$ How can I look at this as being something that I can get through division on the LHS?

Here's a link to a page I found on polynomial long division -- http://www.sosmath.com/algebra/factor/fac01/fac01.html

And again, in terms of step 4 an distributing though, I don't see how I can do this with the missing terms. I know I'm not looking at it correctly as I think that the ellipses means that the starting pattern will go on forever.

No, that's not what the ellipsis (singular, there's just one ellipsis there) means. The ellipsis appears in the middle, so the exponents march down from n-1, n-2, and so on, down to an exponent of 1, and then an exponent of 0.

By distributing in the product $(x - 1)(x^{n -1} + x^{n - 2} + \dots + x + 1)$, what I mean is to multiply x times each term in the right-hand expression, and then multiply -1 times each term in the right-hand expression. You should see that most of the terms drop out.

For example, if I have $(x - 1)(x^3 + x^2 + x + 1)$ I get $x^4 + x^3 + x^2 + x$ + $-x^3 - x^2 - x -1$. If you combine like terms, what do you get?