Yeah, that's not any joke! I'm really gonna prove that!

We know, $2^6=64$

Again, $2^6=(2+0)^6=6_{C_0}2^60^0+6_{C_1}2^50^1+6_{C_2}2^40^2+...=64*0^0$ $⇒64=64*0^0$

Which is a clear proof of $0^0=1$.

True to say, I haven't found any mistake here.

## 0^0=1!

- SINAN EXPERT
**Posts:**38**Joined:**Sat Jan 19, 2019 3:35 pm**Location:**Dhaka, Bangladesh-
**Contact:**

### 0^0=1!

**$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$**

- samiul_samin
**Posts:**1004**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: 0^0=1!

Why $64\times 0^0=64$?SINAN EXPERT wrote: ↑Wed Apr 24, 2019 4:57 pmYeah, that's not any joke! I'm really gonna prove that!

We know, $2^6=64$

Again, $2^6=(2+0)^6=6_{C_0}2^60^0+6_{C_1}2^50^1+6_{C_2}2^40^2+...=64*0^0$ $⇒64=64*0^0$

Which is a clear proof of $0^0=1$.

True to say, I haven't found any mistake here.

- SINAN EXPERT
**Posts:**38**Joined:**Sat Jan 19, 2019 3:35 pm**Location:**Dhaka, Bangladesh-
**Contact:**

### Re: 0^0=1!

$2^6=(2+0)^6⇒64=64*0^0$

**$The$ $only$ $way$ $to$ $learn$ $mathematics$ $is$ $to$ $do$ $mathematics$. $-$ $PAUL$ $HALMOS$**

### Re: 0^0=1!

You know the Binomial theorem:

$ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $

If one of the terms, say $y$, equals 0, then LHS $= (x + 0)^n = x^n$

and, on RHS, the 1st term of the expansion $={n \choose 0} x^n0^0 = x^n0^0$ and all the other terms become 0 because of $0^1, 0^2, 0^3, ... $.

As a good mathematician, you should try to make your theorem or formula as much generalized as possible and make it valid for as many special cases as possible. Now in the case above, $x^n = x^n0^0$, if you define $0^0 = 1$, then your formula extends for the special case of $y = 0$. But if you do not define it that way, your theorem does not work if at least one of the binomial terms is 0. So $0^0$ is a matter of defining, the way we define 0!=1 and similar other terms, and it's a good definition in polynomial algebra.

However, this definition may conflict with your results in Calculus or Limit. So as a good mathematician, you also remind your readers that you say $0^0=1$ for polynomial algebra but undefined in calculus.

You must have noted that as mathematics has advanced and become abstract since its inception, you may not find the exact physical meaning of all mathematical terms. For example, consider even an elementary fraction $\frac{3}{5}$. It's easy to define this fraction with physical objects. You have a loaf of bread, divide it into 5 pieces, and take out 3. But what about $\frac{5}{3}$? You divide the loaf into 3 pieces and take out 5! How could that be possible? But it's again more easily understood if you think of the mixed fraction$1\frac{2}{3}$, although $1\frac{2}{3} = \frac{5}{3}$. So the same thing becomes easy or difficult to comprehend if it is presented in two different mathematical notations.

To summarize, $0^0 = 1$ is by definition in some particular field of algebra. You are right if you agree to that. You are not wrong if you doubt it.

$ (x+y)^n = \sum_{k=0}^n {n \choose k} x^{n - k} y^k $

If one of the terms, say $y$, equals 0, then LHS $= (x + 0)^n = x^n$

and, on RHS, the 1st term of the expansion $={n \choose 0} x^n0^0 = x^n0^0$ and all the other terms become 0 because of $0^1, 0^2, 0^3, ... $.

As a good mathematician, you should try to make your theorem or formula as much generalized as possible and make it valid for as many special cases as possible. Now in the case above, $x^n = x^n0^0$, if you define $0^0 = 1$, then your formula extends for the special case of $y = 0$. But if you do not define it that way, your theorem does not work if at least one of the binomial terms is 0. So $0^0$ is a matter of defining, the way we define 0!=1 and similar other terms, and it's a good definition in polynomial algebra.

However, this definition may conflict with your results in Calculus or Limit. So as a good mathematician, you also remind your readers that you say $0^0=1$ for polynomial algebra but undefined in calculus.

You must have noted that as mathematics has advanced and become abstract since its inception, you may not find the exact physical meaning of all mathematical terms. For example, consider even an elementary fraction $\frac{3}{5}$. It's easy to define this fraction with physical objects. You have a loaf of bread, divide it into 5 pieces, and take out 3. But what about $\frac{5}{3}$? You divide the loaf into 3 pieces and take out 5! How could that be possible? But it's again more easily understood if you think of the mixed fraction$1\frac{2}{3}$, although $1\frac{2}{3} = \frac{5}{3}$. So the same thing becomes easy or difficult to comprehend if it is presented in two different mathematical notations.

To summarize, $0^0 = 1$ is by definition in some particular field of algebra. You are right if you agree to that. You are not wrong if you doubt it.

### Re: 0^0=1!

$0$ শূন্য বহুপদি। শূন্য বহুপদির ঘাত অসজ্ঞায়িত। তাহলে $0$কে তো দ্বিপদি বিস্তৃতিতে ব্যবহার করা যাবে না।

Wãlkîñg, lõvǐñg, $mīlïñg @nd lìvíñg thě Lîfè