I too proved 2=1 , logicaly it cant be - but mathematicaly I made no mistake - can you assure me on that !!?

Question

I came across this question in yahoo answer ::( proved 2=1)
http://answers.yahoo.com/question/index?qid=20070908160205AAjFloI&r=w#HJBWW0fOOE6fdSr17gC3

hence remembered my own acquisition on this matter with different approaches -:

So what or where was my mathematically perceived logic error ?

2 = 1
******
By the common intuitive meaning of multiplication we can see that

4 * 3 = 3 + 3 + 3 + 3

It can also be seen that for a non-zero x
x = 1 + 1 + ........+ 1 ( x terms )

Now we multiply through by x
x^2 = x + x +.............+ x ( x terms )

Then we take the derivative with respect to x
2x = 1 + 1 + ...............+ 1 ( x terms )

Now we see that the right hand side is x which gives us
2x = x

2 = 1

Where is my syntax error ?

kyle x is not a constant, in this question x is assumed an integer; this equation is not meaningful for non-integer real numbers.

kba1a - hey how did you know that

Polexia owns your SO… -: Bring it from the level of the abstract to the level of the concrete; see what happens,. For instance, 1 cup plus 1 cup does not always equal 2 cups. Mix a cup of vinegar with a cup of a baking soda solution. The result will be less than 2 cups of liquid, as some molecules are transformed into carbon dioxide and released into the air as gas.

Jun Agruda, my question (has the meaning of word, thought. Idea, argument, account, reason and principle) its study of valid interference and demonstration in LOGICS is clear. Though its reality and existence might not be a tanible evidence

Jun Agruda, my participation performances here is on the basis of democratic judgment over the suitability of a subject for the intended purposes, as opposed to the authoritarian command, which is meant as an absolute realization of the authority's will, thus not open for debate. Hope you understood what I want you to understand

Answer 1

The problem is in the part you take the derivative.

The derivative of

x + x + . . . + x (x terms).

is NOT

1 + 1 + . . . + 1 (x terms).

Why? Let's review some calculus rules.

D(f + g) = Df + Dg.

I won't prove it but that's true.

From this we can prove (once again proof left as exercise, that)

D( sum from i = 1 to n of f_i)
=
sum from i = 1 to n of (D f_i),

where f_i is a differentiable fucntion.

However, notice that n is a natural number and it's independent of D f_i. You're using a rule that appears to be similar to this but your'e forgetting that the number of times you're summing x by itself is also x, which is itself a variable (not a constant, let alone a constant whole number).

If we neglect what is a variable and what is a constant we run into trouble with things like this:

Notice that the derivative of x^a is

a * x^(a-1).

Notice that the derivative of a^x is

a^x * ln(a).

So is the derivative of x^x

x * x^(x-1) = x^x?

or is it

x^x * ln(x)?

These two rules seem contradictiory. But as it turns out we're not following any of these rules correctly as in each rule a must be a constant value. While we're mixing up a variable x with the constant a.

(In case you're wondering the derivative of x^x is actually

x^x * (1 + ln(x)).

Hope that helps.

Answer 2

Functions are only differentiable on a continuous space such as the real numbers, not on integers. For any particular integer x, you get a true equation. But to differentiate both sides you need an equation of functions, not an equation of integers. The right-hand function x + x + \cdots + x "with x terms" is not a meaningful function on the reals and thus not differentiable.

Also, when taking the derivative in line 4 the derivative is taken with respect to each of the terms individually, but not with respect to the numbers of terms. This is erroneous, as the number of terms is x, the variable of differentiation. The chain rule is incorrectly not applied on the right-hand side of the equation.

Answer 3

I'm pretty sure that I see the problem:

When you expanded x into 1+1+1... (x terms), you failed to acknowledge that the number of terms itself is a function of x. So when you took the derivative, you didn't take that into account. The only way to take it into account is to explicltly say x = x*1... so x^2 = x * x, and so you need to invoke the product rule when differentiating. You've basically "hidden" a factor of x in the RHS.

Note: I've seen the comments on x only being an integer. I'm not absolutely sure that your construction is confined to integers. For example, if x=2.4, then you have:
x = 1+1+0.4 ... and the maths holds.

Answer 4

I'm taking this directly from another topic, which is exactly the same as this one, in which a user named stym (who should be a top contributor in math) made this response:
"x+x+x+x+.........x times ) = x^2
this is true only if x is an integer.
If your domain is integers only, your function is not continuous and you can't differentiate it.
Meaning, you can't go to step 2 of your proof"

Answer 5

You can't take the derivative there, because x^2 = x + x + x... + x only when x is an integer. Under that restriction, the function you are derivating is noncontinuous and thus indifferentiable.

Answer 6

Without logic you won't be able even to use math as a tool. Firstly, you need to have logic. Logic is the science of correct thinking and if you don't have this science how could you proceed with mathematic? So, there's no such thing as logically impossible and mathematically possible. 2 will always be 2 and 1 will always be 1. There's no such thing as 1 = 2 or 2 = 1.

Let's take Bruce Wayne for instance. He is Batman too, isn't it? As regards his being, he is only one. As regards his identities he has two. Here you may say that 1 = 2 but that is a great mistake. He is only one in so far as he exists that is why we say even here that 1 = 1. But in so far as his identities is concerned he has two, referring to them we say 2 = 2. Don't equate things which are dissimilar in nature in order arrive at a fallacy that 2 = 1. This one could accept as true if he has no logic or correct way of thinking. That is why for him here math could be utterly useless. Instead of going towards the truth he may not even be aware that he would be in for more confusion.

Therefore, being is one thing identity is another. Don't get duped you are a philosopher by nature and not a sophos who would want to fool people with their useless, specious and baseless arguments for some reasons or for livelihood or other hidden and treacherous motives of taking advantage of another. Love wisdom and you will see that it won't be a burden to have her and your life will be full of light and joy. You'll be true to yourself and to others and necessarily you will be free.

Answer 7

you're assuming that x is a constant so its derivative would be zero, even the x^2 is a constant so its derivative would also be zero.

Answer 8

You have waaaaaaaaaaaaaaay toooooooooooo much time on your hands. Try taking a little bit of that time and volunteer to a charitable org.

Answer 9

just wondering why it even matters....

think about it in real life. you have two halves of a sandwich, and your friend takes one. then you only have one half.
you had two, and it fills you up.
now you have one, and you're hungry.

haha, im just being annoying, sorry.

Answer 10

you have x raised to the 2nd power (I don't know how you made that sign as an equivalent to 2x in the following line and you can't do that.