Math problems
 

Ok. So this is my first post in the "critical thinking" department so if this is misplaced then sorry. >< I have found two interesting math problems. These are probably old and the answer can probably be found on the internet but I was interested to see what the FFR community would have to offer on these problems:

1. (1/3) + (1/3) + (1/3) = (3/3) That's fine. But now let's use their percent value instead:
(33.3%) + (33.3%) + (33.3%) = (99.9%) {The numbers after the decimals are repeating of course.}
So does this go to show that (3/3) = only 99.9% instead of 100%?
{The same thing can be done to (1/9) or 11.1%. It would work out to be (9/9) = 99.9%.}


2. Let's say A=1 and B=1.
So now we can say that: A=B.
Multiply both sides by A. Now we have: A sqaured = A*B
Subtract (B squared) from both sides: ((A squared) - (B squared)) = AB - (B squared)
Factor: (A + B)(A - B) = B(A - B)
Divide each side by (A - B): (A + B) = B
Substitute: (1 + 1) = 1
Simplify: 2 = 1
If you can't follow this (because I don't know how to put the squared sybol so it may seem confusing) here is a link to a more clear video: here. I read this in my algebra II book last year but didn't think of posting it until now. My teacher never used the book so he never could explain this to the class. But if I remember correctly, the page had this eqaution going down the center and on one of the side notes, it said something like "It seems that 1 = 2. But that is impossible right? So what is wrong with this equation?" Seeming to imply that there is an error in the logic somewhere that I am not able to find.


So I'm interested in seeing how these problems work out. Please post if you have some insight that will clear this up for me. Thanks!

Re: Math problems
 

Interesting. Im not much of a math person...but that makes absolutley no sense.
Perhaps it is that fractions are not a precise way of calculating?

Re: Math problems
 

ok your first problem is a nobrainer 1/3 does equal 33.3% if you are rounding off but the most accurate percent for 1/3 is 33.33333333% now try doing this
33.33333333% + 33.33333333% + 33.33333333%
use a calculator if you have to but you will c that it adds up 2 a 100%

Re: Math problems
 

1-1=0 You can divide by zero in math

Re: Math problems
 

1/3 is a irrational number. That is why you're getting 99.9% 3(1/3)]. It's called a round off error; math isn't perfect.

Re: Math problems
 

Personally, i hate it when people do this.

1/3 does not equal 33.3% its 33.33333 repeating. Once you stop it from repeating, you're rounding the number thus changing the results.
1/3 + 1/3 + 1/3 = 1(100%) fractions have no rounding
.33 + .33 + .33 = .99(.99%) decimals are stopped and rounded

Hope that makes sense

Re: Math problems
 

Fixed

Re: Math problems
 

whoops sorry i meant that lol

Re: Math problems
 

These are all really, really old math problems.

http://www.google.com/search?hl=en&s...+1&btnG=Search
http://www.google.com/search?hl=en&s...em&btnG=Search

And they're not Critical Thinking as they've already been solved. So, moved to Chit Chat.

Re: Math problems
 

Division only makes sense when one of the numbers or factors that you are dividing out is non-zero. You can't divide (A - B) out from each side because (A - B) is zero. Each side can't equal each other because anythng divided by zero is undefined.

Re: Math problems
 

oo and for the 2nd question the reason why 1 = 2 is because of a simple mathematical error it is not possible to divide both sides by a-b since a-b = 0.... in math you cannot divide any number by 0 or it will come out 2 an error

score 2/2 :)

Re: Math problems
 

Ah. You guys are right. I didn't see that I was dividing by 0. I feel dumb now.

I should have seen it. ><
Thanks for pointing this out and thanks for showing me how dumb I am. ><
And sorry for the misplacement. I didn't realize the answers were so easy to get. I thought it would take some actual thought to figure out but I come back 10 minutes later and like 8 people figured them both out.

Edit: Now I feel doubly dumb. ; ; On the video that I posted to make things more clear, like two people pointed out that it was dividing by zero on their comments. The whole second equation could have been left out if I only had read YouTube comments.

Re: Math problems
 

lol yup u feel dumb but u can make up 4 it if you can solve this
X = variable

(1/sin^2 X) + (1/cos^2 X) = (tan X + 1/tan X)^2
no calculators 4 this 1 :D

Re: Math problems
 

Nope. I guess I can't make it up. XD I just finished Algebra II last year and we didn't learn about sine, cosine, and tangent. Well sort of but not how to solve equations with them. I'm about to start Pre-Calculus when school starts on Monday. So I think that I'll be learning that stuff this year maybe.

Re: Math problems
 

wow really?? well anyone can try it if they want then if you guys are still clueless i can give you a hint o n btw i hav no clue wat u ment by alegbra II but i lerned that stuff in a course called functions and relations so i guess its a higher lvl of math then alegbra II

Re: Math problems
 

Well I live in Texas so my Math classes are weird O.o
I'm considered "advanced" so in seventh grade I took Pre-Algebra. Eigth grade was Algebra I. Ninth grade was Geometry. Tenth grade was Algebra II. Now I'm about to be in Pre-Calculus. Then AP Calculus as a senior. If you aren't "advanced" then you take everything a year later than me (you finish off with Pre-Calculus as a senior). I don't know where "functions and relations" rank in my education system.

Re: Math problems
 

lol actually im takin calculus and vectors right now i finished that course a year ago



functions and relations is pre-calculus

Re: Math problems
 

Shame shame, dividing by zero.

Re: Math problems
 

The second statement, I've looked at it and it's interesting. There's also something on Wikipedia I've seen that says zero is equal to one.

Re: Math problems
 

0.99999999999 repeating = 1.

I win. Infinitesimals win. That ****ing dumbass 11th grade math 10th grader at my lunch table who insists I'm wrong about all of this loses.

Re: Math problems
 

Yup. This website may help for anyone who thinks otherwise. Nine different proofs on there.

Re: Math problems
 

We already figured that out but can any1 solve this?

Re: Math problems
 

umm... X = 2 dozen boxes of Pi

lol i could most likely solve it if i was in school, im pretty much shut down even though school started in 4 days

Re: Math problems
 

I'm a little rusty when it comes to trig identities but here goes:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

First I'd combine the first two terms:

(cos^2X+sin^2X)/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

The numerator, cosX^2 + sin^2X, is equal to 1:

1/(sin^2X*cos^2X) = (tan X + 1/tanX)^2

This is the same as:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

So now we are basically trying to equate the things within the second-power-raise, so we aim to equate 1/(sinX*cosX) and tan X + 1/tanX.

Since tan = sin/cos and 1/tan or cot = cos/sin, we see that the right hand side is basically sin/cos + cos/sin, so we have two instances of each. Currently, the left hand side has half as many instances. Therefore we should try to prove this by simplifying the right-hand side:

tanX + 1/tanX
We then change tanX into tan^2X/tanX so we can combine like terms:
(1 + tan^2X)/tanX
And change our tangents into their sine and cosine equivalents:
(1 + (sinX/cosX)^2)/(sinX/cosX)
We then change the 1 into cos^2X/cos^2X so we can combine it as a like term later:
((cos^2X/cos^2X) + (sin^2X/cos^2X))/(sinX/cosX)
Combining terms:
(((cos^2X + sin^2X)/cos^2X))/(sinX/cosX)
cos^2X + sin^2X is equal to 1:
(1/cos^2X)/(sinX/cosX)
Then we simplify things a bit:
cosX/(sinX*cos^2X)
And finally drop the common cosX from the numerator and denominator:
1/(sinX*cosX)

Since we just proved 1/(sinX*cosX) = tanX + 1/tanX, we therefore prove:

(1/(sinX*cosX))^2 = (tan X + 1/tanX)^2

And since we showed that (1/sin^2X) + (1/cos^2X) = (1/(sinX*cosX))^2, we therefore prove:

(1/sin^2X) + (1/cos^2X) = (tan X + 1/tanX)^2

Edit: Oh yeah, QED

Re: Math problems
 

good job rubix you don't look that rusty too me
must have taken a while too get that all down

Re: Math problems
 

For anyone too lazy to understand the (1/3) + (1/3) + (1/3) = 1 thing, I'll write an explanation that hopefully most can understand.

First, 1/3 = 0.333...

The "..." means that the sequence repeats infinitely, to the point where the last digit is a "3" and you cannot add a zero after that. It may sound weird, but trust me, that's the simplest explanation.

Now with simple algebra, we can understand more things.

0.333... = 1/3
0.999... = 1 (Multiply each side by 3)

But how can this be true? This is how:

Oh, and the space filter makes this a little hard to read. Sorry.

Let c = 0.999...

c = 0.999...
10c = 9.999... (Multiply each side by 10)
9c = 9.000... (Subtract 1c from each side)
c = 1.000... (Divide each side by 9)
c = 1 (Simplify)
0.999... = 1 (Substitution property of equality)

Yes, I stole this from Wikipedia.

And about the second problem, it just goes to show you that simplifying after every step is key :D. If you do that, you find out that you get 0=0 early on.