Try to find the error in this proof for 2 = 1
a = b
a^2 = ab
a^2 – b^2 = ab – b^2
(a+b)(a-b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1
Take it to your algebra class and see if you can talk your teacher into extra credit for it.
I know the error!!
It’s…2 = 1
That’s false…
Is it two obvious that you can prove anything you want when you use an invalid common denominator?
True cause for this to be real you would have to divide each side by (a-b) which if a=b that would equal zero. Hey thanks for putting this up here. I just put this in my reply to that other thread. haha. Funny.
-Shaun Johanneson
You can’t divide by 0…so it can’t be real.
Haha, this is a good one, I like it. But yeah, your’s basically going to 1=1 to 0=0, then to 2=1. 0 can be a funny number sometimes.
Man, we unicyclists are a sharp bunch. I’ve run this by some of the engineers I work with and way too many of them forget that division by zero is undefined. Good job.
One other thing, this problem is a unicyclist’s nightmare, cause this means that a bicycle equals a unicycle.
hahahahahahahaha that’s a good one man
i hope this doesnt turn into another infinity thread
dividing by 0 is another form of infinity. Sorry to disappoint you.
sorry to correct, but dividng by zero is not another form of infinity.
the lim as x approaches zero of (lets say) 1/x is infinity(of course we have to discuss the direction we are approaching from(neg versus pos inf.))
but we cannot divide by zero! period. it makes no conceptual sense.
fun problem, ill ask my alg kiddos later in the year to find the error
beat me to it!
Haha, I know that. You can’t perform the operation of a number over 0, but you need limits as it aproaches, I just worded it a little too simply. but anyway, the discussion of infinity is on the other math thread, and maybe we should try to keep it there, and not discuss i here.
You guys I just figured this out now. 1+2=3, 3+4=7, 7+8=15, 1+5=6, 6+7=13,
1+3=4, 4+5=9, 9+10=19, 1+9=10, 1+0=1, and it starts over. look at the pattern. Equations until it breaks up again, 3-2-3-1-4-2-3-1-4 repeating. Look at this 32314231423142314 (2314)
2314 = in letters bcad, mixing them around you get bad C, and in my experiences a C for a grade was bad. Creepy huh?
-Shaun Johanneson
um… i dont know what to say
Certainly not a coincidence. I think you may be onto something there…
Instead of posting another thread, I’ll put it here.
123,456,789 x 8 + 9 = 987,654,321
That other one is awesome!
Ed
Here’s a tricky math problem that a lot of people probably have heard of.
Three guys stay in a hotel for one night, and need to pay for the room. The fee is $30, so each of the fork out $10 a piece. One of them goes down to pay the person working, but tell’s the guy that there’s a $5 discount for that night. Rather than try to split up the five dollars evenly among the three men, the guy decides to leave a two dollar tip, then split up the three ones, this way it’s much easier. This way, each guy gave the hotel $9 a piece for the room. Let’s take a closer look:
$9 x 3 men=$27
$27 + $2 tip = $29
Where is the missing dollar?
That’s as much of a math problem as a Rubik’s Cube is a topology problem.
It’s more puzzle than math.