A nice math riddle, try to solve it.

[Lior]

Hello...
Believe it or not, but it seems like I can prove you that 2=1.
Look how:

a=b /Multiply both sides by a
a²=ab /Add to both sides (a²-2ab)
a²+a²-2ab=ab+a²-2ab
2a²-2ab=a²-ab
2(a²-ab)=a²-ab
2=1

Of course there is a mistake here, find it.

Good Luck!

[Good Dreams]

2(a²-ab)=a²-ab

Since a=b we can do this:

2(a²-a²)=a²-a²

that is:

2 *0=0 <=> 0=0

I don't know if that error as a name of some sort though.
Hmm, I just thought I bit more. You are dividing something by zero, which is always risky.
Actually it's zero dividing by zero (my brain is slow in the morning), which is in fact undefined.

Post more ridles about this, it's fun!

[Lior]

Very good.

You solved the riddle.

We devided by (a²-ab) and since a=b its like (a²-a²) which is 0.
And as we know, deviding by zero is forbidden because it is undefined.

[Good Dreams]

Originally posted by Lior
And as we know, deviding by zero is forbidden because it is undefined.

What? We can't say for example that 9/0=0? I only thought it was undefined when we divided 0 by 0.

[parksie]

0/0 = 1

x/0 = undefined where x != 0

[Good Dreams]

Error in my last post!

I meant:

Can't we say that x/0 is infinite where x<>0?
I thought only when 0/0 is undifined.

0/0 isn't 1 , is it????

[parksie]

Infinite == Undefined (== should be = with 3 lines [identity])

[Lior]

Guys, its pretty ironic that I have to say this, but a basic math rule is that we cant devide ANY number by 0.
deviding by 0 is undefined!

this includes 0/0 or 9/0 or whatever.

Anything cannot be devided by 0.

[parksie]

It depends whether you apply

x/0 is undefined

before

x/x = 1

[Lior]

Since 0/0 is undefined, so a/a=1 was applied AFTER the first rule.

I mean, mathematicians said a/a=1 except in one case when a=0.

You surely dont think they said "a/0=undefined, oh, by the way, a/a=1". Do you?

[parksie]

Well, we learnt a/a = 1 first

But I see your point

[Guv]

Girsl & guys, a good reason for 0/0 not being allowed or defined is the proof that started this post.

If 0/0 = 1, then the proof that 2 = 1 is valid.

BTW: I was taught a simpler variation of this paradoxical proof. It went as follows.

a^2 - b^2 = (a + b) * (a - b)

For a = b, the above results in the following.

a^2 - a * b = (a + b) * (a - b)

a* (a - b) = (a +b ) * (a - b)

divide by (a - b) to get

a = a + b, where a = b. Hence a = 2 * a or 1 = 2

[Olly]

It's so obviously! We don't know the values af a or b. So don't know whether a=b, that's what we assume. As 2 is not equal to 1, a is not equal to b, because if a would equal b, then we would get the same values on each site as we added the same to each. so the whole thing is not more than a blow hahaha

[Good Dreams]

Did you read the whole post? a=b is assumed from the beginning,

[Olly]

we would get 2=1 if we wouldn't know that a=b.
but we know it, so a gets devided by b as b =a, as you already said it
but very interesting, sorry for that *blow* hehe

[Lior]

Originally posted by Olly
we would get 2=1 if we wouldn't know that a=b.

Let me correct you:

We would NOT get 2=1 if we wouldn't know that a=b.
Or simpler: we will get that 2=1 if we know that a=b.

[Olly]

After your logic we could proof that a=b because 2=1, that's ridiculous. If it would be so, then the whole mathematics would be wrong and as they aren't your calculation was wrong.

[Zaei]

Logically, 0/0 should be 0, since 0 parts of 0 apples is still 0 parts... -.-

Z.

[Lior]

If so, 0 / 0 should be 1, as a / a = 1.

[Zaei]

Perhaps 0/0 is a special case, with more than one solution (like square roots), where it could be 0 and it could be 1.

Z.

[Guv]

Girls & guys, 0 / 0 being undefined is a well established concept. Division is defined as the inverse of multiplication, although it does not have to be as long as the definitions are consistent with the following.

A * B = C implies C / A = B and vice versa.

Applying the above to 0 / 0 = X implies that X * 0 = 0

Any finite value of X will satisfy X * 0 = 0 implying that any finite value can be assigned to the quotient 0/0. Hence 0/0 has no specific defined value.

In some circumstances, 0 / 0 is assigned a value to be consistent with previous analysis. Consider the following trivial situation.

Assume that Y = (3 * X^2 + B) / (4 * X^2 + B) has been shown to be the result of some analysis.

For B = 0 and X = 0, Y = 0 / 0, which in this case would be treated as 3 /4.

This done because for B = 0, Y = 3 / 4 for all values other than X = 0 It would seem strange to claim that when B = 0, then Y = 3 / 4 for all values of X, except for X = zero, in which case it might be something else.

It is also defined that way for more complex reasons, namely the following.

If Y = FunctionA( X) / FunctionB( X), and both functions equal zero (or infinity) for some value of X, Y is defined as DerivativeA( X ) / DerivativeB( X ).

[Arbiter]

No, no, NO!!!

b = b + (a*2)

where

a = a cheescake
b = my belly

[VictorB212]

If I remember correctly

where x != 0

x/0 is undefined

0/0 is worse than undefined, it tells you nothing.

If you remember back to working on rational functions (polynomial over polynomial), if you got 0/0 you knew nothing about the point. However, if you get x/0, you know there's an asymptote.

For example:

x^2 +  x - 6
------------
x^2 + 2x - 3

Try the following numbers:
{-3,2,1}

-3 : yields 0/0, which tells you nothing about what exists at x = -3
2 : yields 0/-5, or a valid 0
1 : yields -4/0, or undefined -- this tells you that there is a verticle asymptote here

0/0 is more than undefined, it's more like nothingness

[Guv]

VictorB212: 0! has been covered in another Thread. As for the other issues, I believe that your analysis is faulty. These issues are subtle. I believe that the following is generally considered valid by mathematicians.

Division by zero is viewed as an undefined operation, and explict expressions like 1/0 & 0/0 are avoided like the plague. Instead, mathematicians work with the behavior of functions as variables approach certain values.

Mathematicians often deal x = 0 in functions like y = 1/x by saying that as x approaches zero, y grows without bound. The claim that y = infinity when x = 0 is generally considered jargon or a short way of saying that y grows without bound as x approaches zero. The value of y when x equals zero is usually considered undefined or not to be considered usable.

Functions which can result in 0/0 for certain values of a variable are analyzed by considereing the behavior of the function as the variable approaches the critical value. In the case of your function, the factor of (x+3) would be factored out and the function would be viewed as (x-2)/(x-1), which is 1.25 when x = -3.

Mathematicians would claim that 1.25 is the value of your function when x = -3. A more sophisticated approach would yield the same result. Without factoring out (X+3), it would be noted that the function reduces to 0/0 when x = 3. To analyze further, the derivative of the numerator and the derivative of the denominator would be analyzed, resulting in consideration of the following.

(2*x+1)/(2*x+2), which is -5/-4 when x = -3, resulting in a value of 1.25 as above.

There is some theorem which states that certain quotients can be assigned the value of the quotient of derivatives.

Actually, I think that mathematicians hedge a bit by stating in the above case that no contradictions arise from assigning 1.25 to the function at the critical value of x.

Please note a previous post about A/B = C implying B*C = A and vice versa. Since X*0 = 0 for any finite value of X, then 0/0 can be assigned any finite value without introducing some paradox.

[VictorB212]

Guv

"!=" does not stand for factorial. It stands for the logical NOT operator.

I wasn't trying to get into any depth. I was merely stating that 0/0 tells you less than x/0 where x != 0 (that means x does not equal 0). Is that argument faulty? If so, then my math teacher is wrong

Oh well.

[Guv]

VictorB212: Your teacher is not telling you the whole story.

If a mathematician were to analyze your example function, he would not state that for X = -3 the function is undefined, even though substituting -3 for X would result in 0/0.

He would note that the function is equal to (X-2)/(X-1) for every every value of X other than X = -3, and conclude that it should be defined as equal to (X-2)/(X-1) for X = -3 as well as for oll the other values. Hence he would conclude that the function was equal to 1.25 when X = -3.

I would hesitate to say that 0/0 is worse than 1/0, because 0/0 can be assigned a finite value usable for further computations, while 1/0 cannot be coped with. After all if 5*0 = 0, why not 0/0 = 5? Of course 6*0 = 0, hence 0/0 = 6 is not so bad either.

I repeat, mathematicians deal with the behavior of functions as variables approach some critical value. They avoid discussion of expressions like 0/0 and 1/0 when isolated from a context.

When being formal, mathematicians avoid the common notion that 1/0 equals infinity. They prefer making statements like "The function 1/X grows without bound and X approaches zero. "