Teacher teasers... (1 Viewer)

[biggie walls]

My teacher has brought up the following two things... but has said she cannot teach it to us because it will take too long and is off the track...

1) dividing by zero x/0

2) and i dont know what she meant from this but she said in uni she had to derive equations from 'dot'... as in ---> .

Have a crack someone??
It would be really interesting to see these phenomena realised...

(oh I do 2 unit btw so you'll prob have to explain it in lamens terms)

 

[Mongke]

As for your second question, I cannot attempt to answer it as I don't understand what it means.

correct me if im wrong here, but thats the whole point!

 

[Riviet]

How does the twelve line proof go like? I'm just curious to know.

 

[acmilan]

the only thing i can think of for the 'dot' thing is dot product, which is a sort of multiplication, except multiplying vectors, not numbers.

 

[Rax]

Well we have discussed a mild teaser with our teacher once

As you know any number to the power of 0 is =1

BUT what about 0^0 (zero to the power of zero)

That can't equal 1.....can it?
You smart people answer me now
lol

 

[A l]

Zero to the power of zero would be undefined because you'd be dividing by zero. In other words to find say 2º (without knowledge of the result of course), you would multiply 2ª by 2(-ª) because when you add the indices you get an index of zero. Since anything multiplied by its reciprocal is 1, then any real number to the power of zero should be 1. However, in the case of zero, to obtain it, you would multiply 0ª by 0(-ª), BUT 0(-ª) cannot be defined because you cannot divide by zero, therfore 0º is undefined. If 0ª were to ever equal 1, then 0/0 would equal 1 (which is considered indeterminate).

 

[acmilan]

LOL 0⁰ resulted in a very, very, very long debate between myself, Xayma, who_loves_maths and brett. It lies somewhere in this forum, in a thread started by dreamerish i think.

 

[Templar]

Division by 0 is not allowed because when you divide, you're actually multiplying by the inverse. 0 doesn't have an inverse, so you can't divide by it.

As for 1+1=2, I'm aware of two main proofs. The 1+1=1'=2 by Peano's axioms, and the 400 page one constructing from Zermelo Fraenkel.

 

[Sober]

Iruka, your proof is flawed, apply it to 0².

0² = 0^3-1= 0³ / 0¹ = 0 / 0 = indeterminate.

There is a reason that it is called indeterminate rather than undefined, it means it may have a value but is impossible to solve without extra information.

Interestingly Google calculator seems to disagree with us all.

 

[icycloud]

As Iruka noted, although 0^0 is indeterminable, one can find the limit of x^x as x approaches 0.

Let u = x^x, lnu = xlnx

Now
lim{x-->0+} lnu
= lim{x-->0+}xlnx
= lim{x-->0+} lnx/(1/x)
= Infinity/Infinity {Thus, L'Hopital's Rule can be applied}

Therefore,
lim {x-->0+} lnu
= lim{x-->0+} lnx/(1/x)
= lim{x-->0+} (1/x)/(-1/x^2) {By L'Hopital's Rule}
= lim{x-->0+} -x
= 0

Thus,
lim{x-->0+}u
= lim{x-->0+}x^x
= e^0
= 1

That is to say, x^x = 1 as x approaches 0.

L'Hopital's rule is a handy little rule. For example, you can use it to show lim{x-->0} sinx/x = 1 (though not in the HSC). As such:

lim {x-->0} sinx/x = 0/0 {Indeterminate and thus L'Hopital's Rule is applicable}
lim {x-->0} sinx/x
= lim{x-->0} cosx/1 {By L'Hopital's Rule}
= cos 0
= 1

For more info on the rule: http://mathworld.wolfram.com/LHospitalsRule.html

 

[Rax]

LOL Oh God I hope I havent started up the 0^0 thing again

I read through that Old Thread lolololololol

It was pretty good for the discussion

Who Loves Maths really went at it eh........You people are so too smart

So dont start it up again, I dont want my head on the block lol

GG