# How 3=2 is true?,help me to find out the mistakes

• Nero26
In summary: It's all very confusing. But anyway, back to the conversation at hand:In summary, the conversation discusses the use of derivatives to solve the identity of m^2+m^2+m^2+...up to m term = m^2*m = m^3, and the potential mistakes that may arise in the process. One potential mistake is the different orders of operation used in differentiating the left and right hand sides. Another possible mistake is considering the number of terms as a variable when differentiating the right hand side, which leads to m^3 instead of 2m^2. Ultimately, the conversation concludes that taking the derivative of a sum with respect to
Nero26
m^2+m^2+m^2+...up to m term =m^2*m=m^3
Which is identity .So we use derivative for this identity.
Now, d/dm(m^2+m^2+m^2+...up to m term)=d/dm(m^3)
Or 2m+2m+2m+...up to m term=3m^2
Or 2m*m=3m^2
Or 2m^2=3m^2 but it is not possible.So there should be some mistakes.But I'm not getting it.
What I suspect are as follows:
In L.H.S we differentiated first,then add all terms.In R.S we added all then differentiated it.Is the mistake for maintaining different orders of operation?
I suspect another reason like while differentiating R.S, number of terms is considered as variable and multiplied with m^2 that leads to m^3,but as it is no. of terms if I assume it as constant then, m*d/dm(m^2)=m*2m=2m^2,which matches with L.S. Is this okay?or something else?

Last edited:
Here is my thought:
you have m-terms of m^2. That is, m*m^2. Taking derivative, d\dm (m*m^2)=m^2+m*2m=3m^2.
The problem with d/dm(m^2+m^2+m^2+...up to m term) is that m-term is also a function of m. So, technically, you need to differentiate the function (m-term)

Nero26 said:
m^2+m^2+m^2+...up to m term =m^2*m=m^3
Which is identity .So we use derivative for this identity.
Now, d/dm(m^2+m^2+m^2+...up to m term)=d/dm(m^3)
Or 2m+2m+2m+...up to m term=3m^2
Or 2m*m=3m^2
Or 2m^2=3m^2 but it is not possible.So there should be some mistakes.But I'm not getting it.
What I suspect are as follows:
In L.H.S we differentiated first,then add all terms.In R.S we added all then differentiated it.Is the mistake for maintaining different orders of operation?
I suspect another reason like while differentiating R.S, number of terms is considered as variable and multiplied with m^2 that leads to m^3,but as it is no. of terms if I assume it as constant then, m*d/dm(m^2)=m*2m=2m^2,which matches with L.S. Is this okay?or something else?

You are not allowed to take a derivative of a sum like ##S(m) = \sum_{i=1}^m a_i## with respect to m. Think about it: if S(m) had a derivative, it would be
$$S'(m) = \lim_{h \to 0} \frac{\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i}{h} = \lim_{h \to 0} \frac{\sum_{i=m}^{m+h} a_i}{h}.$$ If you take, for example, h = 0.001, what on Earth could you possibly mean by ##\sum_{i=m}^{m + 0.001} a_i?## It makes no sense.

Thanks to all for your help.
Ray Vickson said:
You are not allowed to take a derivative of a sum like ##S(m) = \sum_{i=1}^m a_i## with respect to m. Think about it: if S(m) had a derivative, it would be
$$S'(m) = \lim_{h \to 0} \frac{\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i}{h} = \lim_{h \to 0} \frac{\sum_{i=m}^{m+h} a_i}{h}.$$ If you take, for example, h = 0.001, what on Earth could you possibly mean by ##\sum_{i=m}^{m + 0.001} a_i?## It makes no sense.
I couldn't understand this line,
##\sum_{i=1}^{m+h} a_i - \sum_{i=1}^m a_i=\sum_{i=m}^{m+h} a_i##
##\sum_{i=m}^{m+h} a_i=a_m+a_{m+1}+a_{m+2}+...+a_{m+h}## (assuming h is some integer)But terms upto ##a_m## should be canceled by ##\sum_{i=1}^{i=m}a_i## then what remains is ##a_{m+1}+a_{m+2}+...+a_{m+h}## which is ##\sum_{i=m+1}^{i=m+h}a_i## .Now in this case as h→0, after subtraction we'll get a term ##a_{m+h}## that fails to exist.
Another thing I couldn't understand is:
$$S(m)=\sum^{i=1}_{i=m}a_i$$,Whether a is constant or function of m?
However ,it seems I got your point that as number of terms can't be a fraction so we can't get derivative,like ##a_{m+0.001}## can't exist.Am I right?

Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.

R: How 3=2 is true?,help me to find out the mistakes

Your left hand term is only defined for integer values of m, so d/dm makes no sense.

LCKurtz said:
Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.
Yes, order has been restored. We can all sleep easy tonight!

LCKurtz said:
Whew! Glad to see that's settled, because it is well known that 3 = 2 only for a large value of 2.
Or a very small value of 3...

## 1. How can 3 ever equal 2?

The statement "3=2" is a mathematical equation, not a literal statement. In mathematics, the equal sign (=) means that the expression on the left side is equivalent to the expression on the right side. Therefore, in this equation, 3 and 2 are considered equal because they represent the same value.

## 2. What mistakes could lead to the misconception that 3 equals 2?

In most cases, the mistake is usually a result of a mathematical error, such as incorrect calculation or a misplaced decimal point. It could also be due to a misunderstanding of mathematical concepts, such as the meaning of the equal sign or the difference between integers and fractions.

## 3. How do you explain the concept of "3=2" to someone who doesn't understand math?

When explaining this concept, it's essential to first clarify that in mathematics, the equal sign does not mean "is the same as." Instead, it indicates that both sides of the equation have an equivalent value. So, in the case of "3=2," it means that 3 and 2 are equivalent in this specific equation.

## 4. Can 3 ever equal 2 in a real-life scenario?

No, in the real world, 3 and 2 are distinct and cannot be considered equal. However, the equation "3=2" is an abstract representation of a mathematical concept and does not necessarily reflect real-life situations. In mathematics, we often use symbols and equations to simplify complex concepts and make them easier to understand and manipulate.

## 5. What is the significance of understanding the concept of "3=2"?

Understanding the concept of "3=2" is crucial in building a strong foundation in mathematics. It helps us understand the meaning of the equal sign and how mathematical equations work. Furthermore, it allows us to solve more complex equations and apply mathematical concepts in various fields, such as science, engineering, and finance.

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