Wednesday, 11 July 2012

0^0 ( Zero Raised To The Zeroth Power ) Equals ?

What does 0^0 (zero raised to the zeroth power) equal?

0/0 = ?

Here comes some ANSWERS from different talented students , few teachers and also from Mathematician .

Clever Student :

I know it .......... !!

$x^{0}$ = $x^{1-1}$ = $x^{1} x^{-1}$ = $\frac{x}{x}$ = $1$ .

Now we just plug in x=0, and we see that zero to the zero is one!

Cleverer Student:

No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:

$0^{x}$ = $0^{1+x-1}$ = $0^{1} \times 0^{x-1}$ = $0 \times 0^{x-1}$ = $0$

which is true since anything times 0 is 0. That means that

$0^{0}$ = $0$ .

Cleverest student :

That doesn’t work either, because if $x=0$ then

$0^{x-1}$ is $0^{-1} = \frac{1}{0}$

so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function $x^{x}$ and see what happens as x>0 gets small. We have:

$\lim_{x \to 0^{+}} x^{x}$ = $\lim_{x \to 0^{+}} \exp(\log(x^{x}))$

= $\lim_{x \to 0^{+}} \exp(x \log(x))$

= $\exp( \lim_{x \to 0^{+} } x \log(x) )$

= $\exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )$

= $\exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )$

= $\exp( \lim_{x \to 0^{+} } -x )$

= $\exp( 0)$

= $1$

So, since $\lim_{x \to 0^{+}} x^{x}$ = 1, that means that $0^{0}$ = 1.

High School Teacher:

Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$ . The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

Calculus Teacher:

For all $x>0$ , we have

$0^{x} = 0$ .

Hence,

$\lim_{x \to 0^{+}} 0^{x} = 0$

That is, as x gets arbitrarily close to $0$ (but remains positive), $0^{x}$ stays at $0$ .

On the other hand, for real numbers y such that $y \ne 0$ , we have that

$y^{0} = 1$ .

Hence,

$\lim_{y \to 0} y^{0} = 1$

That is, as y gets arbitrarily close to $0$ , $y^{0}$ stays at $1$ .

Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$ . In particular, when we approach (0,0) along the line with x=0 we get

$\lim_{y \to 0} f(0,y) = 1$

but when we approach (0,0) along the line segment with y=0 and x>0 we get

$\lim_{x \to 0^{+}} f(x,0) = 0$ .

Therefore, the value of $\lim_{(x,y) \to (0,0)} y^{x}$ is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$ .

Mathematician:

Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.

Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:

$y^x$ := $1 \times y \times y \cdots \times y$

where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get

$y^{x}$ = $1 \times y$ .

However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving

$y^{0}$ = $1$

which holds for any y. Hence, when y is zero, we have

$0^0 = 1$ .

Look, we’ve just proved that $0^0 = 1$ ! But this is only for one possible definition of $y^x$ . What if we used another definition? For example, suppose that we decide to define $y^x$ as

$y^x$ := $\lim_{z \to x^{+}} y^{z}$ .

In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.

[Clarification: a reader asked how it is possible that we can use $y^z$ in our definition of $y^x$ , which seems to be recursive. The reason it is okay is because we are working here only with $z>0$ , and everyone agrees about what $y^z$ equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]

Interestingly, using this definition, we would have

$0^0$ = $\lim_{x \to 0^{+}} 0^{x}$ = $\lim_{x \to 0^{+}} 0$ = $0$

Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$ . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.

So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for x>0 and y>0 we know what we mean by $y^x$ . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.

But if this is the case, then how can mathematicians claim that $0^0=1$ ? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined. For example, consider thebinomial theorem, which says that:

$(a+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$

where $\binom{x}{k}$ means the binomial coefficients.

Now, setting a=0 on both sides and assuming $b \ne 0$ we get

$b^x$

= $(0+b)^x$ = $\sum_{k=0}^{\infty} \binom{x}{k} 0^k b^{x-k}$

= $\binom{x}{0} 0^0 b^{x} + \binom{x}{1} 0^1 b^{x-1} + \binom{x}{2} 0^2 b^{x-2} + \hdots$

= $\binom{x}{0} 0^0 b^{x}$

= $0^0 b^{x}$

where, I’ve used that $0^k = 0$ for k>0, and that $\binom{x}{0} = 1$ . Now, it so happens that the right hand side has the magical factor $0^0$ . Hence, if we do not use $0^0 = 1$ then the binomial theorem (as written) does not hold when a=0 because then $b^x$ does not equal $0^0 b^{x}$ .

If mathematicians were to use $0^0 = 0$ , or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using $0^0 = 1$ .

There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

Tuesday, 10 July 2012

0/0 Equals Two (Fallacy)

0/0 Equals Two (Fallacy)

0/0 = (100-100) / (100-100)
= (10-10)(10+10) / 10(10-10)
= (10+10) / 10
= 20 / 10
= 2
0/0 = 2

Thus 0/0 = 2 is proved by the help of Mathematical Fallacy .

Monday, 9 July 2012

Mystery Math On 0/0

Mystery Math On 0

In Mathematics most of the operations have been made easy using few tricks or even by normal methods . But many of us gets a great confusion in getting value this expression :

0/0 = ?

Each person gives each value for this expression but none can go with single value with confidence.

In some point of view , the value of this expression is 1 , as they compare this expression with the expression ,
  x/x = 1 ( A Number Divided By Itself Is ONE )

In other views , the value will be taken as 0 on comparing this expression with ,

0/x = 0 ( ZERO Divided By Any Number Is ZERO )

Few consider the value of this expression as INFINITY , due to the expression ,

x/0 = INFINITY
( Any Number Divided By ZERO Is INFINITY )

Thus there are THREE values considered for this expression - 1 , 0 , INFINITY. So which one of these values is to be considered as the exact value for this . We cannot choose any value as we wish and also it is impossible to go on with one value . Upon all these conditions , 0/0 is considered as an UNDEFINED FUNCTION or UNDEFINED VALUE .

Saturday, 7 July 2012

Impossible Mind Reader - Card Trick

Impossible Mind Reader - Card Trick

Remove 25 cards from the pack and allow the spectator to shuffle and cut the cards as much as they want then lay the 25 cards out in 5 rows of 5 cards face upwards from left to right in neat rows following a sort of grid pattern.

Ask the spectator to think of any one of the cards in any row and to tell you which row their card is located in. When they tell you, note the card at the left hand end of the row. Take up the cards, beginning first at the last or right hand card of the bottom line, placing it on the face of the card immediately above it and then place these 2 cards on top of the card immediately above, and so on up to the top.

Do this with each of the rows in turn and you should be left with 5 piles of cards on what was the top row.

Collect up the cards, placing the 5th (right hand) pile on top of the 4th, then this stack on top of the 3rd, then on top of the 2nd, etc, until you're left with a single stack of cards.

Then once again, keeping the cards face up, deal then into another 5 horizontal rows as before. Ask the spectator to tell you once again which row their card is located in. When they tell you, look along the top or the bottom row for the card which you noted earlier was first in the row they had indicated. Above it or below in the row to which they now point, is the card which they originally thought of.

Monday, 18 June 2012

Etruscan Numerals

Etruscan Numerals

The Etruscan numerals were used by the ancient Etruscans. The system was adapted from the Greek Attic numerals and formed the inspiration for the later Roman numerals.There is very little surviving evidence of these numerals. Examples are known of the symbols for larger numbers, but it is unknown which symbol represents which number. The general argueement among the Etruscologists is :

Etruscan	Decimal
θu	1
zal	2
ci	3
śa	4
maχ	5
huθ	6
semφ	7
*cezp	8
nurφ	9
śar	10
*θuśar	11
*zalśar	12
*ciśar	13
huθzar	14
*maχśar	15
*śaśar	16
ciem zaθrum	17
eslem zaθrum	18
θunem zaθrum	19
zaθrum	20
cealχ	30
*huθalχ	40
muvalχ	50
śealχ	60
semφalχ	70
cezpalχ	80
*nurφalχ	90