A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Quick Question: Exponents I know this has been asked so many times but what is the proper explanation to why \(\sf n^0=1\) and \(\sf 0^0=0\) ?

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh i get the first one.. from the law of exponents \(\sf \Large \frac{a^m}{a^n}=a^{m-n}\) If \(\sf m=n\), it will be like: \(\sf \Large \frac{a^m}{a^m}=a^{m-m}=a^0\) for example, let's say \(\sf 2^3\) \(\sf \Large \frac{2^3}{2^3}=2^{3-3}=2^0\) this is the same as\(\sf \frac{8}{8}=1\) But how about the second one.. 0^0 is equal to zero and not 1?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is it because any number you multiply by zero is still zero or something like that lol 0/0 is not usually "undefined" from what i know zero is an interesting number hmm

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's not true that \(0^0=0\). Consider this rewriting: \[0^0=e^{\ln0^0}=e^{0\ln0}\] Let's say for the moment that \(0\ln0\) makes sense. Then that would mean that there's some number \(n\) such that \(e^n=0\). But there is no such \(n\).

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, that's the real point there. so if 0^0 = 0 is not true, is it okay to say that 0^0 = 1?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That wouldn't work either. Consider \(f(x)=x^0\) and \(g(x)=0^x\). What happens as \(x\to0\) for each?

  6. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    0^0 is undefined just as 1/0 is. It makes no sense. x^0 = 1 for x!=0 for notation purposes :)

  7. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    or you could look at it the other way, properties of exponents are the reason...

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x^0 will always be in y=1, while the other one will be at y=0. I think I got it, but I just don't know how to explain it. This probably means that 0^0 doesn't have a "right" value for it. I just asked this because I saw an article which says that they set 0^0=1 as a standard answer for 0^0 :P

  9. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    just as you would explain 1/0. It actually makes no mathematical sense.

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a proof using properties of exponents? can you show it to me?

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I suppose it can be convenient at times to define \(0^0\) to take on the value of 1. It depends on the context, in the same way that defining \(\dbinom nk=0\) for \(n<k\) might help in smoothing out some calculation.

  12. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(0^0=1\\0*ln(0)=0\) yikes!

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The point is that the limits of \(0^x\) and \(x^0\) as \(x\to0\) suggest that \(0^0\) could actually take on two values, but that would break some axiom (possibly the reflexive? I forget the names; the one that says \(x=x\) or something to that effect).

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The key word here is "indeterminate". The so-called number can behave one way from one perspective, and behave in a completely different way from another. Another indeterminate form that might be easier to grasp is \(\dfrac{0}{0}\). If we suppose there is some number \(x\) for which \(x=\dfrac{0}{0}\), then that would suggest that \(0x=0\), which is true for all \(x\). But we start off by assuming \(x\) has a definitive value. Yet \(0x=0\) is true for all \(x\). We can't *determine* its value, hence the name "indeterminate".

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    assume 0^0 is well defined. then by properties of exponents 0^0= 0^(3-3) = 0^3/0^3 = 0^3* 1/0^3 = 0*1/0 but 1 / 0 is undefined contradiction

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep, got it! I think I'm satisfied with all your explanations. I hope I can give you all a medal, but I can't so I'll just fan you. Thank you guys for all your useful inputs :)

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just thought of this. I'm not sure if it's an airtight proof. Seems logical to me :)

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.