You Can't Divide By Zero!

Number Theory Level 1

$\Large \frac{0!+0!+0!+0!}{0!+0!}= \ ?$

0 1

\infty

12 solutions

Sudoku Subbu
Jan 29, 2015

The main issue here is to determine $0!.$ We should know from the definition of factorials that $0!=1.$

Thus, $\frac{0!+0!+0!+0!}{0!+0!}=\frac{1+1+1+1}{1+1}=\frac{4}{2}=2.$

But you're welcome to share how 0!=1 .

It'll be helpful for those beginners in Combinatorics.

Muhammad Arifur Rahman - 6 years, 4 months ago

n!=(n+1)!/(n+1)

So, 0!=1!/1=1

Archit Boobna - 6 years, 4 months ago

Thanx for telling... :)

Rishabh Tripathi - 6 years, 3 months ago

Wow I never knew that n! = (n+1)!/(n+1) ... so just a quick question a bit off topic ... if I understood correctly does that mean that 0! and 1! are equal ? Because as you wrote 0! = 1!/0=1 and 1! = 2!/2 = 1 ? Thanks for the education today by the way !

Mellisa Vezina - 5 years, 11 months ago

@Mellisa Vezina – Yes 0! And 1! Are equal

Saurab Thakur - 5 years, 11 months ago

@Mellisa Vezina – Correct it 0!=1!/1

Shiv Kumar - 5 years, 10 months ago

If 0!=1 and 1!=1 then 0!=1! Therefore 0=1 how?

Omkar Chavan - 5 years, 7 months ago

@Omkar Chavan – Let's see if this helps. You could see factorial as a function, and not every function have a different output value for every different input value, as long as no two input values are the same when given an output value. One example would be f(x)=1. In this case, f(0) and f(1) (or in fact any input value) will result in a same output value.

Margaret Zheng - 5 years, 4 months ago

@Margaret Zheng – Also, n! is supposed to be how many ways you can arrange n objects, with all n objects distinct and thus since there is 1 way to arrange 0 objects, just do nothing, 0!=1 and 1!=1 since you can only arrange 1 object in 1 way, just leave it there.

Razzi Masroor - 4 years, 4 months ago

Good, you're smarter than Sudoku Subbu .

Muhammad Arifur Rahman - 6 years, 4 months ago

Here is my note on 0!=1

Yoogottam Khandelwal - 5 years, 11 months ago

Your note is amazing, just like your Avatar Picture! Thank you so much for your Note!!!

Muhammad Arifur Rahman - 5 years, 11 months ago

@Muhammad Arifur Rahman – Thanks : ) I love Harry Potter

Yoogottam Khandelwal - 5 years, 11 months ago

yes i will?!?! we know that $n!=\frac{(n+1)!}{n+1}$ so when we substitute 0 in place of n, $0!=\frac{(0+1)!}{0+1}=\frac{1}{1}=1$ hence proved! if you need proof even for $n!=\frac{(n+1)!}{n+1}$ i am free to say it

sudoku subbu - 5 years, 11 months ago

Its been a it late for you to reply. But finally you've saved the new learners. That's appreciating.

Best wishes!

Muhammad Arifur Rahman - 5 years, 10 months ago

2 is correct answer thank you

asad khan - 5 years, 11 months ago

Can we apply firstly the bodmas rule?

Amit Sharma - 5 years, 11 months ago

Yes, you can. But then you will take more time to solve then. that'l make the cake a hard chips!

Muhammad Arifur Rahman - 5 years, 10 months ago

Nicholas Tanvis
Oct 20, 2015

Actually, it is pretty easy. You do not even need to know what 0! is. $\frac{0!+0!+0!+0!}{0!+0!}=\frac{4(0!)}{2(0!)}=2$

You would need to know that 0! ≠ 0, surely?

Mark Grindley - 2 years, 11 months ago

Hadia Qadir
Sep 7, 2015

0 != 1
so = (1+1+1+1)/(1+1)
= 4/2 = 2

Its not 0=1, Its 0!= 1

Correct it, Qudir.

Muhammad Arifur Rahman - 5 years, 9 months ago

thanks ! writing mistake

Hadia Qadir - 5 years, 9 months ago

Now your solution deserves an UPVOTE. So I've voted your solution.

Write more and more solutions on Brilliant. Make Brilliant better.

Muhammad Arifur Rahman - 5 years, 9 months ago

@Muhammad Arifur Rahman – yes I do . thank you yet again

Hadia Qadir - 5 years, 9 months ago

Raj Kumar
Jul 1, 2015

I took a different approach to the problem.

Suppose, x = 0!

Then problem becomes:

$\frac{x+x+x+x}{x+x}$ = $\frac{4x}{2x}$ = 2

But I believe Lu Chee Ket and Sudoku Subbu's approach is much authentic.

how if x=0?

Calvin Wong - 5 years, 11 months ago

Incorrect substitution.

This should be it.

$\frac { 0!+0!+0!+0! }{ 0!+0! } \\ Let\quad x=0,\\ \frac { x!+x!+x!+x! }{ x!+x! } \\ =\frac { 4x! }{ 2x! } \\ =\frac { 4 }{ 2 } \\ =2$

Edit: According to the statement $x!,x\ge 0$ .

Kenneth Choo - 5 years, 11 months ago

Good, you've corrected Raj Kumar's Solution.

Muhammad Arifur Rahman - 5 years, 11 months ago

Why x = 0! is a wrong substitution? Is there any mathematical background to it?

Raj Kumar - 5 years, 11 months ago

If we take 0!=1 Then firstly we have to solve the division portion By BODMAS Ans is 5

Amit Sharma - 5 years, 11 months ago

Both are incorrect substitutions you cant substitute x=0 correct substitution X=0+h (lim h->0) it becomes 4x!/2x! =2x!/x! Applying L hospital rule (d/dx (2x!))/ d/dx (x!) =2/1 =2

Prakhar Gupta - 5 years, 11 months ago

I admit I don't understand your solution since I haven't learnt that yet. However, I do realize I missed out a statement.

$x!,x\ge 0$

And $0$ is within that range. With this, I'm sure my solution is correct and of course, your solution can be correct too. But still, I prefer the method of showing

$0!=1!=1$

and then substituting $1$ into $0!$ .

Kenneth Choo - 5 years, 11 months ago

@Kenneth Choo – L hospital rule is no statement its a way of solving limits it involes differential calculas i m not just canceling 'x' it is being removed by differentiation

The problem with x=0 is that simple algebra fails as 0/0 is indeterminate form type 0/0 on google This is the reply "In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (assuming a≠0), and so division by zero is undefined. Since any number multiplied by zero is zero, the expression 0/0 also has no defined value and is called an indeterminate form."

Prakhar Gupta - 5 years, 11 months ago