The image size limit has been raised to 1mb! Anything larger than that should be linked to. This is a HARD limit, please do not abuse it.
Our new Indie Games subforum is now open for business in G&T. Go and check it out, you might land a code for a free game. If you're developing an indie game and want to post about it, follow these directions. If you don't, he'll break your legs! Hahaha! Seriously though.
Our rules have been updated and given their own forum. Go and look at them! They are nice, and there may be new ones that you didn't know about! Hooray for rules! Hooray for The System! Hooray for Conforming!

Maths

Registered User, __BANNED USERS regular
edited January 2011
Just don't understand the mystery rule here, I got the asnwer in the back of the book I just don't know why they did it like this.

1+1/x
1/x

Why doesn't the 1/x just cancel out the top one to make it one? I guess the book wants me to do x/1 (1/x) but I am not sure why you don't cancel it out. Obviously both can't work since the former gives me 1 and the latter 1+x

Fizban140 on

Posts

• When the last moon is cast over the last star of morning And the future has past without even a last desperate warningRegistered User, Moderator mod
edited January 2011
The bottom one can't cancel the top one because the top one is being added and is not a factor. You can only cancel factors.

ceres on
And it seems like all is dying, and would leave the world to mourn
Dear satan

• NYCRegistered User regular
edited January 2011
1 + 1/x is a term
If you're talking about canceling something the entire term must possess the element you want to cancel. So in this case you would only be able to cancel out the 1/x if the top term was 1/x(1+1/x) or 1/x + (1/x)*(1/x)

minirhyder on
• Registered User regular
edited January 2011
you can't cancel it out because that operation you wrote there is essentially adding fractions, not necessarily just dividing.

Basically -> [1 + (1/x) ] / (1/x) = [1 / (1/x) ] + [(1/x) / (1/x)]

Does that make sense?

from here you can see that [(1/x) / (1/x)] = 1 and you can figure out that [1 / (1/x) ] = x

and thus this = x+1

does this help at all?

Al_wat on
PSN: AWATTT66| XBox Live: AWATTT66| Steam: AL-WAT| Battle.Net: ALWATTS #1320
Origin: aiwatt
• Registered User regular
edited January 2011
like heres an example that might be easier to read.

A + B
C

this expression is the same as

A + B
--- ----
C C

To us an example that is more of the form of your question; it would be like this:

A + C
C
=

A + C
--- ----
C C

Al_wat on
PSN: AWATTT66| XBox Live: AWATTT66| Steam: AL-WAT| Battle.Net: ALWATTS #1320
Origin: aiwatt
• NYCRegistered User regular
edited January 2011
I'll try again Al_wat's terminology:

A + B
B

You cannot cancel out because both of the terms in the numerator must have the term you're trying to cancel out; these are just the rules of math.
Think of it as doing the opposite of the distributive property [ie A(B+C) = AB + AC], that is, in order to cancel out an element from the term, every single element of that term must have the element which you are trying to factor out or cancel out. For instance if you had

AB + CB
B

You can cancel out the B and end up with A + B
You can also think of this as factoring B out of the expression and turning it into B(A+C)

However in the example you have and in this example

A + B
B

You cannot do so because the entire term does not have B in it. If you were to try to factor out B from the term A + B you would be unable to do so, and that is your problem.

minirhyder on
• Registered User regular
edited January 2011
Here's the full solution, if you're interested in figuring out exactly what happens in it. It's got some pretty important algebra rules in it.

finalflight89 on
• Registered User regular
edited January 2011
That's an awfully convoluted solution - the inverting fraction knowledge used in the second last step could have been used at step one (ie, if you know the dividing by a fraction is the same as multiplying by an inverse, you can just do that straight off and get x * (1 + 1/x) as step 2).

The key thing to know here is that if you multiply the numerator and the denominator by the same thing, the fraction remains unchanged (because x/x = 1 and multiplying by 1 changes nothing)

(1 + 1/x) / (1/x)

= (x * (1 + 1/x)) / (x * (1/x))

= (x + 1) / (1)

= x + 1

soxbox on
• Registered User, __BANNED USERS regular
edited January 2011
soxbox wrote: »
That's an awfully convoluted solution - the inverting fraction knowledge used in the second last step could have been used at step one (ie, if you know the dividing by a fraction is the same as multiplying by an inverse, you can just do that straight off and get x * (1 + 1/x) as step 2).

The key thing to know here is that if you multiply the numerator and the denominator by the same thing, the fraction remains unchanged (because x/x = 1 and multiplying by 1 changes nothing)

(1 + 1/x) / (1/x)

= (x * (1 + 1/x)) / (x * (1/x))

= (x + 1) / (1)

= x + 1
That makes a lot of sense, thanks for the responses everyone.

Fizban140 on