Daedalus

when we are using any calculating device which is of ten (10) digits than its largest positive number is (+L) +9999999999 and smallest negative number is (-L) -9999999999

The above assumption is incorrect. The largest / smallest possible number for any electronic calculating device is not [math]\pm 9999999999[/math]. It is actually based on the the precision of the FPU as described here:

 

Double-precision floating-point format

 

Furthermore, you have neglected to consider calculators that are based on arbitrary precision arithmetic.

 

In computer science, arbitrary-precision arithmetic, also called bignum arithmetic, multiple precision arithmetic, or sometimes infinite-precision arithmetic, indicates that calculations are performed on numbers which digits of precision are limited only by the available memory of the host system.

You have erroneously related the base of the number system to the number of digits for your largest / smallest number. My TI-89 can handle base 10 numbers above and below your specified range.

 

Now if there is one fraction X divided by Y (X/Y) and Y ≠ ±L than for that fraction two fractions whose values will be nearest to it will be X divided by largest number preceding Y (X/Y') and X divided by smallest number succeeding Y ( X/Y'').

In this situation X/( Y')> X/Y>X/( Y'') and all three will be approximately equidistant from each other.

Here you are trying to apply the squeeze theorem. However, there is a huge problem with your method. You do not apply the limit correctly. Let me demonstrate the issue. You are claiming that if:

 

[math]z = \frac{x}{y}[/math] then [math]z = \frac{x}{2}\left(\frac{1}{y'}+\frac{1}{y''}\right)[/math] where

 

[math]y' = \lim_{n \to 0^-} (y-n)[/math] and [math]y'' = \lim_{n \to 0^+} (y+n)[/math] such that

 

[math]\frac{x}{y'} > \frac{x}{y} > \frac{x}{y''}[/math].

 

However, as stated in your method, you are not taking the limit as [math]n[/math] approaches zero. Instead, you are claiming that calculating devices cannot cover the entire set of reals, which makes it possible to choose some value [math]\epsilon > 0[/math] such that:

 

[math]z = \frac{x}{2}\left(\frac{1}{y'}+\frac{1}{y''}\right)=\frac{x}{y}[/math]

 

The above statement is wrong because we are dealing with the set of real numbers!!! What you are actually doing is as follows:

 

[math]z = \lim_{n \to \epsilon} \frac{x}{2}\left(\frac{1}{y-n}+\frac{1}{y+n}\right)=\frac{x\, y}{y^2-\epsilon^2} \ne \frac{x}{y}[/math]

 

Example 1 - you claim that:

 

A = X/2 [ 1/Y' + 1/Y'' ]

 

Here, X= 100, Y'= 4.999999999 and Y''= 5.000000001

 

Therefore, A = 100/2 [ 1/4.999999999 + 1/5.000000001 ]

 

[since we are using ten digits calculating device therefore we can use only ten digits. Therefore using ten digits [to] approximate [the] value of both we get]

 

= 50 [ 0.2 + 0.2 ]

= 50 [ 0.4 ]

= 20

 

...

 

Since we can see that denominators Y' and Y'' are not having their prime factors as only two (2) and five (5) therefore using L.C.M. technique we get,

 

A = 100/2 [ (5.000000001+ 4.999999999)/((4.999999999)× (5.000000001) ) ]

= 100/2 [ 10/((5 - 0.000000001)× (5 + 0.000000001) ) ]

 

Using identity of (a – b) (a + b) = a^2 - b^2 we get,

 

A = 100/2 [ 10/(5^2 –〖 (0.000000001)〗^2 ) ]

= 100/2 [ 10/(25 – 0.000000000000000001) ]

= 100/2 [ 10/24.999999999999999999 ]

 

Since we are using ten digits calculating device therefore we can use only ten digits. Therefore using ten digits approximate value we get,

 

A = 100/2 [ 10/25 ]

= 100/5

= 20

The problem with the above example is that:

 

[math]\left(\frac{1}{5-0.000000001}+\frac{1}{5+0.000000001}\right) \ne \frac{2}{5}[/math]

 

Your method only approximates the result of the division:

 

[math]\left(\frac{1}{5-0.000000001}+\frac{1}{5+0.000000001}\right) \approx \frac{2}{5}[/math]

 

You cannot claim that this allows you to divide by zero over the set of reals. It is a logical fallacy:

 

[math]\left(\frac{1}{0-0.000000001}+\frac{1}{0+0.000000001}\right) \ne \frac{2}{0}[/math]

 

Now I will address your latest post:

 

1. You are taking B.N.R.F. in the aspects of differential and integral calculus, the domain of mathematics where non-uniform velocity (displacement/time) can also be treated as uniform velocity by using the concepts of differentiation and integration. Which in real life doesn't exists at all but in mathematical and scientific aspects it is believed to be happening. However B.N.R.F. deals with the real life aspects of division. The divisional aspects which takes place in our daily life or/and in our calculating device. Because the calculating device is nothing but the subset of real number line.

Your method is based on limits, which forces us to evaluate it with the tools of calculus. The above statement is pure speculation because you are trying to relate your method to a physical application of dividing by zero. We are not talking about physical processes (although it would not matter if we were because the mathematics of indeterminate forms - [math]1/0, 0/0[/math], etc... - can have limits that are zero, infinity, etc...). We are discussing the result of a division by zero as defined by your method:

 

Conclusion

We can see that both these results obtained by Bhartiya New Rule for Fraction completely resembles with the results stated by Brahmagupta on 628 A.D. and Mahavira on 830 A.D. and this only enables us to establishment of Universal Results for Division By Zero.

 

1. According to Brahmagupta Zero divided by Zero is Zero and furthermore any positive or negative number divided by Zero is a fraction with that number on numerator and Zero as denominator from which we can conclude that he had stated that any number divided by Zero is Zero.

2. According to Mahavira the number remains unchanged when divided by Zero i.e. when any number is divided by Zero than it gives us the same number as remainder.

3. According to Bhaskara II any number divided by Zero tends to infinity (∞).

4. According to Bhartiya New Rule for Fraction any number divided by Zero give us quotient as Zero and that number as remainder.

 

On combining all these four above statements we get ;

 

According to Bhartiya New Rule for Fraction;

The value of fraction any positive or negative number say X divided by Y tends infinity as the value of Y tends to Zero.

However when any number say X is ultimately divided by Zero it gives Zero as a quotient and that number as a remainder.

 

This is where you reveal that you are not truly dividing by zero. Defining a subset of real numbers does not allow for division by zero:

 

For example: Two sets associated with Bhartiya New Rule for Fraction when we are using ten (10) digits calculating device are:

 

A. Set of real (Actual) number line. Set A = { X ; X ∈ R }

B. Set of calculating device number line. Set B = { X ; -L ≤ X ≤ +L and X ∈ 10 digits number }

 

From the two sets above it is clear that set B is subset of set A. Therefore if we apply any function on that Set B or derive any formula for them than it will be automatically applicable in all those elements of real number line which are in that subset B.

 

At this point once again I like to clear the fact that we are only studying the property of Set B (Set of calculating device number line) to generalize and obtain some concrete results on the aspects of division by Zero.

 

2. Now, Second thing, when we are using any calculating device which is of ten (10) digits than its largest positive number is (+L) +9999999999 and smallest negative number is (-L) -9999999999. Now if there is one fraction X divided by Y (X/Y) and Y ≠ ±L than for that fraction two fractions whose values will be nearest to it will be X divided by largest number preceding Y (X/Y') and X divided by smallest number succeeding Y ( X/Y'').

 

...

 

Once again have a look over my point of view that, we are only studying the property of Set B (Set of calculating device number line) to generalize and obtain some concrete results on the aspects of division by Zero. And the above result is only for the particular calculating device.

 

So, in this case there will be no any equation in three variables as mentioned in your last post and it will simply be equal to X/Y for that particular subset of real number line.

The problem with this statement is that it does not matter which set you choose, A or B. We are not studying the properties of set B. We are analyzing the operations that you are applying to the set. Once you realize this, you will see that your method is not dividing by zero:

 

[math]\frac{x \times 0}{0-\epsilon^2} \ne \frac{x}{0}[/math]

 

It does not matter if we write your method as an equation of two variables or three. The results are the same:

 

[math]\frac{x \, y}{(y-0.000000001)(y+0.000000001)} \ne \frac{x}{y}[/math]

 

or

 

[math]\frac{x \, y}{(y-\epsilon)(y+\epsilon)} = \frac{x \, y}{y^2-\epsilon^2} \ne \frac{x}{y}[/math]

 

The only reason why your method produces a zero when [math]y = 0[/math] is because:

 

[math]\frac{x \times 0}{0 - \epsilon^2} = \frac{0}{-\epsilon^2} = 0[/math]

 

 

3. While using Bhartiya New Rule for Fraction we are calculating only up to limited number of digits because of problem of non-termination of division. According to modern Mathematics If denominator of any fraction not having any common factor other than one (1) has its prime factors as only 2 and 5 than it have a terminating value which reveals that the value of it is a perfect number, whereas if the same has its prime factor as other numbers than that of only 2 and 5 than it have a non-terminating value. In such situations we use their approximate values and calculating device too shows their values with appropriate approximation. (Article from Mystery of Zero - Shoonya Ka Rahasya, Chapter 3, Page no.31)

The key point that I will make here is that you are calculating up to a limited number of digits. This means that your method has two discontinuities where [math]y^2 = \epsilon^2[/math]. However, because you state that you are only considering set B, your method cannot approximate values of [math]|y| > L[/math] because you incorrectly suggest that a base 10 calculating device cannot handle numbers larger or smaller than [math]\pm L[/math].

 

There is one more thing I would like to add in response to your claims:

 

The interesting points (or claims being made) in regard of this formula are:

 

1. 'Bhartiya New Rule for Fraction' is based on present phenomenon and rules of mathematics.

 

2. It is very simple and easy formula.

 

3. Greatest benefit of 'Bhartiya New Rule for Fraction' is that it is capable of dividing by Zero and giving its value as an integer.

 

4. 'Bhartiya New Rull for Fraction' can be used to find out the value of four not defined trigonometric ratios tan 90, cosec 0, sec 90, cot 0. So that these values can be utilized in the field of astronomy and other fields related to mathematics.

 

5. If in place of simple division (x/y), 'Bhartiya New Rule for Fraction' is used in any digital electronic device as its processing command for division in processor, it will results in permanently elimination of 'divide by Zero' error from that device.

1.) Your method is not based on the rules of mathematics for division. It is based on the function [math]\text{B}(x,y)=\frac{x\, y}{y^2-\epsilon^2}[/math] that you have defined where [math]x[/math] and [math]y[/math] are variables, and [math]\epsilon[/math] is a constant such that [math]\epsilon > 0[/math].

 

2.) Stating that your method is simple and easy to use is an opinion and not a fact.

 

3.) Your method when [math]y=0[/math] is equivalent to [math]\frac{x \times 0}{0-\epsilon^2} = -\frac{0}{\epsilon^2} \ne \frac{x}{0}[/math] and does not divide by zero. However, [math]f(x, 0)=0, \ \ \{x \in \mathbb{R} \ | \ -\infty < x < \infty\}[/math].

 

4.) Your method produces incorrect values for the ratios tan 90, csc 0, sec 90, and cot 0.

 

Proof for cot 0:

 

 

Trigonometric functions:

 

[math]\text{tan} \ \angle A = \frac{a}{b}, \ \ \ \ \text{cot} \ \angle A = \frac{b}{a}[/math]

 

Bhartiya New Rule for Fraction:

 

[math]\text{B}(x, y) = \frac{x \, y}{y^2 - \epsilon^2}[/math] where [math]\epsilon[/math] is a constant such that [math]\epsilon > 0[/math]

 

The slope of a line is equal to the change in y divided by the change in x, which corresponds to the definition:

 

[math]m=\frac{\Delta y}{\Delta x}=\text{tan} \ \angle A[/math] where [math]\angle A[/math] is the angle of the line with respect to the [math]x[/math] axis.

 

It is known that two lines, [math]L_1[/math] and [math]L_2[/math], are perpendicular if their slopes [math]m_1 \times m_2 = -1[/math]. If [math]L_1[/math] has a slope of [math]m_1=\text{tan} \ \angle A[/math], then a line [math]L_2[/math] perpendicular to [math]L_1[/math] will have a slope [math]m_2=-\text{cot} \ \angle A[/math]:

 

[math]m_1 \times m_2=(\text{tan} \ \angle A) \times (-\text{cot} \ \angle A) = -1[/math]

Now consider a horizontal line that has a slope of zero [math]m_1=0[/math]. The line perpendicular to it has an undefined slope [math]m_2=\infty[/math]. However, if we replace division with [math]\text{B}(x,y)[/math], the affected tan and cot functions produce:

 

[math]m_1 = \text{tan} \ 0 = 0[/math]

[math]m_2 = -\text{cot} \ 0 = 0[/math]

 

This results in two horizontal lines that are parallel to each other:

 

[math]y_1=0 \times x + b_1[/math]

[math]y_2 = 0 \times x +b_2[/math]

 

You have you successfully broken trigonometry and linear algebra.

As for claim 5, you are more than welcome to modify your calculating devices to use your method instead of actually doing division. Just leave my devices alone : ) In closing, you should consider the following:

 

[math]\lim_{y \to 0} \frac{0}{y} = 0[/math]

 

which is similar to:

 

[math]\lim_{y \to 0} \frac{x \, y}{y^2 - \epsilon^2} = 0[/math]

 

However, these limits are not equal:

 

[math]\lim_{y \to 0^-} \frac{1}{y} = -\infty \ \ \ \ \lim_{y \to 0^+} \frac{1}{y} = \infty[/math]

This is why we call these things indeterminate forms.

The following graph allows us to visualize the problem:

 

 

[math]\vec A[/math] is the position vector for where the ball was kicked, [math]\vec v_b[/math] is the velocity vector of the ball, [math]\vec B[/math] is the position vector where the player starts running north, [math]\vec v_p[/math] is that player's velocity vector, and [math]\vec C[/math] is the position vector where the player intercepts the ball.

 

You already know parts a - c:

 

a.) The speed of the ball is equal to the magnitude of [math]\vec v_b[/math] such that:

 

[math]|\vec v_b|=\sqrt{5^2+8^2}=\sqrt{89}\, m/s[/math].

 

b.) The position vector of the ball after [math]t[/math] seconds is:

 

[math]r_b(t)=\vec A + t \, \vec v_b=(2+5t)i+(1+8t)j[/math].

 

c.) The time when the ball is due north from [math]\vec B[/math] is when the [math]i[/math] component of the ball's position vector equals the [math]i[/math] component of [math]\vec B[/math]:

 

[math]2+5t = 10\, m, \ \ t=8/5 \, s = 1.6\, s[/math]

 

Now for part d:

 

1.) You already know it takes [math]1.6\, s[/math] for the ball to be located due north of [math]\vec B[/math].

2.) You also know the position that the player at [math]\vec B[/math] begins running for the ball.

3.) You also know that the player runs due north such that the position vector of the player after [math]t[/math] seconds is:

 

[math]r_p(t)=\vec B + t \, \vec v_p=10i+(7+v\, t)j[/math]

 

Now all you need to do is set the position vector of the ball [math]r_b(t)[/math] equal to the position vector of the player [math]r_p(t)[/math] at time [math]1.6\, s[/math] and solve for [math]v[/math]:

 

[math](2+5(1.6))i+(1+8(1.6))j = 10i+(7+v(1.6))[/math]

 

[math](2+5(1.6))i = 10i[/math]

[math](1+8(1.6))j = 13.8j = (7+v(1.6))j[/math]

 

The rest is up to you : )

You still haven't answered these questions Ankur! Division by zero is undefined by definition for good reason. Why should we throw good reason out with the bath water for an unneeded, irrational solution?

doG, I wouldn't pay much attention to what he is selling. I have already shown that he is confused about dealing with indeterminate forms. He is comparing a function of three variables with a function of two, and trying to say that both are equal to each other.

Derivation of Bhartiya New Rule for Fraction

 

To understand derivation of Bhartiya New Rule for Fraction it is important to know some facts like, between any two numbers there are infinite numbers therefore, it is impossible to say which number (Y') is the largest number which precedes any number (Y). Similarly, it is also impossible to say which number (Y'') is the smallest number which succeeds any number (Y). ...

 

Just because you are taking a limit does not mean that you are dividing by zero. You are applying Calculus to analyze the following equation:

 

[math]\frac{x \, y}{y^2 - z^2}[/math]

 

which is not equal to:

 

[math]\frac{x}{y}[/math]

 

I would recommend that you learn some Calculus before making such assertions. You are incorrectly comparing two different indeterminate forms.

I came across a formula derived by Ankur Tiwari, which he says enables division by zero.

 

The website claims

 

This formula enables us to divide in a unique way without using denominator. This formula is based on the principle that, If the value of X divided by Y (X/Y) is A than by using this formula we can find out A without dividing X by Y directly, that means without dividing X by Y we can find out its value. This is the reason why 'Bhartiya New Rule for Fraction' is capable of diving by Zero.

 

The interesting points in regard of this formula are :-

 

1.'Bhartiya New Rule for Fraction' is based on present phenomenon and rules of mathematics.

 

2. It is very simple and easy formula.

 

3.Greatest benefit of 'Bhartiya New Rule for Fraction' is that it is capable of dividing by Zero and giving its value as an integer.

 

4.'Bhartiya New Rule for Fraction' can be used to find out the value of four not defined trigonometric ratios tan90, cosec0, sec90, cot0. So that these values can be utilized in the field of astronomy and other fields related to mathematics.

 

5.If in place of simple division (X/Y), 'Bhartiya New Rule for Fraction' is used in any digital electronic device as its processing command for division in processor, it will results in permanently elimination of 'divide by Zero' error from that device.

 

How would it affect mathematics? Is there a fallacy in there?

Let us discuss...

 

As Cap'n Refsmmat pointed out:

 

The above formula approximates X / Y by taking the average of the division at points to the left and to the right of Y. It's a very informal limit.

This actually implies that the Bhartiya New Rule for Fraction is ill defined, and does not actually equal the result of the division. This mathematically flawed rule is based on the following limit:

 

[math]\lim_{n \to 0}\frac{x}{2} \left(\frac{1}{y-n}+\frac{1}{y+n}\right)=\lim_{n \to 0}\frac{x}{2} \left(\frac{y+n}{(y-n)(y+n)}+\frac{y-n}{(y+n)(y-n)}\right)=\lim_{n \to 0}\frac{x}{2} \left(\frac{2y}{y^2-n^2}\right)=\frac{2 \, x \, y}{2 \, y^2}=\frac{x}{y}[/math]

 

However, B.N.R.F. is based on the precision of a floating point processor (FPU). Therefore, [math]n[/math] does not approach 0, but some other value [math]\epsilon[/math]. This affects the result of the limit:

 

[math]\lim_{n \to \epsilon}\frac{x \, y}{y^2-n^2}=\frac{x \, y}{y^2-\epsilon^2}=\frac{x}{y}\left(\frac{1}{1-\frac{\epsilon^2}{y^2}}\right) \ne \frac{x}{y}[/math]

 

The reason why this method produces incorrect results when dealing with division by zero is as follows:

 

[math]\lim_{y \to 0}\frac{x \, y}{y^2-\epsilon^2}=\frac{0}{0-\epsilon^2}=0, \ \left \{\epsilon \in \mathbb{R} \, | \, \epsilon \ne 0 \right \}[/math]

 

Where the actual limits for [math]\frac{x}{y}[/math] as [math]y[/math] approaches zero are:

 

[math]\lim_{y \to 0^-}\frac{x}{y}=-\infty[/math]

 

and

 

[math]\lim_{y \to 0^+}\frac{x}{y}=\infty[/math]

a 4 part equalateral pyramid of a trigonomic equation that would in use of a cofficent be rendered to sphere with the latitude being formed to the relation of pi. then deflated by using the same method then the arc of pi is captured in a 4 part equation to the trigonomic form. then the use of pythagorus theorem is used in a trigonomic form in the negitative space for a elipical co efficent then rendering what is similar to a atom structure that would soon bond to another gravitational pull of another atom. This would mean that the equation that would render to the form that would be pi in a constant state of flux would render to a geographical location of a digit to a mathmathical formula. This in essence means that quantum physics is no longer a science.

With the negitative space that is one in this structure the structure of the universe as the relativity time and space light speed formula is put in place to the effect of the pi and the time and space rendered to a model in application to the formula would end up with what is a order of all creation.

 

As a king would be represented in this equation that would be the star of David.

 

The potent of all creation to come like a ballon to all things of the creator spoken to existence, is also the blackhole explained in the consuming of that creation to what is said in revelations as be hot or cold if your lukewarm you will be spat out. Every action has a reaction. Yet it further states in the word, that God is the center meaning that the very center of creation is the center to the break down of what we see as a chemistry distilation flask. In 3 parts God. Father, Son, Holy spirit. To what we see as the basic of the properties on this earth. Plasma requires the solid, liquid, and gas state together to make such.

After reading your word salad sprinkled with terminology obtained from watching the Discovery Science channel and Back to the Future, I will have to refer you to rule 1 of the speculations forum:

 

Speculations must be backed up by evidence or some sort of proof. If your speculation is untestable, or you don't give us evidence (or a prediction that is testable), your thread will be moved to the Trash Can. If you expect any scientific input, you need to provide a case that science can measure.

 

I have to agree with Ophiolite:

 

This is bringing down the tone of the Speculations forum - not an easy thing to do.

Perhaps the Trash Can would be a better forum for this thread.

Hey guys,

 

http://www.sciencefo...tics-tutorials/

 

the presentation of the thread is good indeed. it help me a lot to get some intuitive on learning calculus, but only 1 thread? well, currently I'm struggling with integrals; as far as I understand it, it's said to be "opposite of differentiation". yet, like "differentiation" topic, I can't grasp on how the symbol works. Furthermore, there so lack of resources on integration topic on the net.

 

I am planning to do some tutorials on integration once this semester is over. I've already been working with Capn' on this, and you can see that the last posting in the tutorials section was contributed by me. I just haven't had time lately to finish my tutorial on Riemann sums. I plan to not only cover integration, but also give real world examples of how to apply the integral.

This may be of some interest to the discussion:

 

Galaxy Rotation Curves from General Relativity with Renormalization Group Corrections

 

We consider the application of quantum corrections computed using renormalization group arguments in the astrophysical domain and show that, for the most natural interpretation of the renormalization group scale parameter, a gravitational coupling parameter G varying [math]10^{-7}[/math] of its value across a galaxy (which is roughly a variation of [math]10^{-12}[/math] per light-year) is sufficient to generate galaxy rotation curves in agreement with the observations. The quality of the resulting fit is similar to the Isothermal profile quality once both the shape of the rotation curve and the mass-to-light ratios are considered for evaluation. In order to perform the analysis, we use recent high quality data from nine regular disk galaxies. For the sake of comparison, the same set of data is modeled also for the Modified Newtonian Dynamics (MOND) and for the recently proposed Scalar Tensor Vector Gravity (STVG). At face value, the model based on quantum corrections clearly leads to better fits than these two alternative theories.

Damn those lying dutchmen... I guess it's up to me to make a video of my sons and I flying with wings made of wax.

You and your kids would probably end up flying too close to the sun, they'd melt and you'd be in the ocean....

 

Poor Icarus...

I definitely want wings like this:

 

The numbers remind me of this:

 

It's genius though! It's no wonder numbers have always boggled scientists! We've always been trying to find big numbers using just digits from 1 to 9. No one ever thought of using a 14 before!

 

 

Troll though? Maybe we'll never know. It's not like one could just check the other videos uploaded by the same user on youtube.

I think it's great. From "it's too much of a taboo subject" and "People couldn't work out... squares" to the closing inspirational thoughts, this is just enough like other serious crackpot theories that it's treated like one.

 

Ok... I watched the video and others by the OP. Let's just say that I'm going to be nice and withhold my comments.

Besides, I have discoverd my own equations which can produce numbers larger than yours:

 

http://www.sciencefo...es-by-daedalus/

 

Oh... you are also not allowed to steal my equations. However, feel free to use them. Just make sure you bring them back when you are done.

I could care less about your calculations. Prime numbers are amongst the biggest known numbers that have meaning behind them. They are also quite common in science and mathematics. We still have no clue what your number means, and I don't think most of us actually care. It's not like we would win a Fields Medal for knowing your method.

...is this guy for real?

 

I'm not sure. He may be a troll from Narnia for all I know.

Here is that link you missed:

 

http://primes.utm.edu/largest.html

http://primes.utm.ed.../page.php?id=14

 

Ooo.... this one is huge:

 

http://www.math.utah...largeprime.html

 

I can keep this up all day. If you are interested in large numbers than visit this site:

 

http://www.mersenne.org/

 

They love to prove that these large numbers are primes.

Well, if the numbers I have listed are bigger than your number, then clearly your number is NOT the biggest number of all time. Oh... prime numbers are probably more proper than yours.

Not a proper number? You just made 7,212,610,147,295,474,909,544,523,785,043,492,409,969,382,148,186,765,460,082,500,085,393,519,556,525,921,455,588,705,423,020,751,421 cry :'( The answer to what? The meaning of life?

 

This prime number is bigger and has 100 digits:

 

7,212,610,147,295,474,909,544,523,785,043,492,409,969,382,148,186,765,460,082,500,085,393,519,556,525,921,455,588,705,423,020,751,421

 

This prime number is even bigger yet with 200 digits:

 

40,992,408,416,096,028,179,761,232,532,587,525,402,909,285,099,086,220,133,403,920,525,409,552,083,528,606,215,439,915,948,260,875,718,893,

797,824,735,118,621,138,192,569,490,840,098,061,133,066,650,255,608,065,609,253,901,288,801,302,035,441,884,878,187,944,219,033

 

This prime number is bigger than your number and it's a palindrome:

 

742,950,290,870,000,078,092,059,247

 

"12978189 digits"

 

12978189 is a smaller number then 7,987,987,778,981,078,910,148

 

That's just the count of the digits making up the number and not the number itself. Your number only has 22 digits.

Have you ever heard of the Ackermann function? The numbers produced by this function can easily become larger than your 22 digit number. Also, a googolplex [math]10^{10^{100}}[/math] is by far larger than your number and only uses a one and a whole bunch of zeroes. We even know about prime numbers larger than your number:

 

http://primes.utm.edu/largest.html

 

The largest known prime from 2008 is [math]2^{43112609}-1[/math] and has 12978189 digits.

When one is unable to distinguish between the possibilty of skillfull troll or the mentally ill it is arguably time to close a forum thread.

 

I second that notion.

Yes you can actually solve this problem without knowing the radius of the cylinder. Since I can't actually give you the answer, I will help you set up the equations so you can solve it. First, let's define some variables that will make our equations clear and understandable.

 

The height of the cylinder is:

 

[math]H_{cyl} = 6.70 \, \text{cm}[/math]

 

The height of the cylinder that is submerged or under water is:

 

[math]H_{sub} = 6.70 \, \text{cm} - 1.70 \, \text{cm} = 5.0 \, \text{cm}[/math]

 

You should know that the volume of a cylinder is:

 

[math]V = h\, \pi \, r^2[/math]

 

Now that we have the above variables and equation for the volume of a cylinder, we can begin to solve this problem.

 

The magnitude of buoyancy force may be appreciated a bit more from the following argument. Consider any object of arbitrary shape and volume V surrounded by a liquid. The force the liquid exerts on an object within the liquid is equal to the weight of the liquid with a volume equal to that of the object. This force is applied in a direction opposite to gravitational force, that is of magnitude:

 

[math]B=\rho_f \, V_{disp} \, g[/math]

 

where ρ_f is the density of the fluid, V_disp is the volume of the displaced body of liquid, and g is the gravitational acceleration at the location in question.

 

If this volume of liquid is replaced by a solid body of exactly the same shape, the force the liquid exerts on it must be exactly the same as above. In other words the "buoyancy force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to:

 

[math]B=\rho_f \, V \, g[/math]

We can see from the above quote from Wikipedia that the magnitude of the buoyancy force is:

 

[math]B=\rho_f \, V_{disp} \, g[/math]

 

We know that the force of gravity is trying to pull the cylinder down and that the buoyancy force is trying to raise the cylinder up. The force of gravity acting upon the cylinder is:

 

[math]F = m\, g[/math]

 

However, the cylinder is neither sinking or rising. This means that the net force on the cylinder is zero and the magnitude of the buoyancy force is equal to the magnitude of the gravitational force:

 

[math]F_{net} = m\, g - \rho_f \, V_{disp} \, g = 0[/math]

 

We can rearrange the terms to show that the forces are equal in magnitude:

 

[math]m\, g =\rho_f \, V_{disp} \, g[/math]

 

Now that the forces are balanced, we can divide both sides by the gravitational acceleration to obtain the mass of the cylinder:

 

[math]m =\rho_f \, V_{disp}[/math]

 

 

 

You should be able to work the problem from here on because:

 

1.) You know that the density of water is [math]\rho_f = 1\, \text{gram} / \text{cm}^3[/math]

 

2.) You know that the volume of water displaced is equal to the volume of the cylinder that is submerged [math]V_{disp} = H_{sub}\, \pi \, r^2[/math]

 

3.) You know that the mass of the cylinder is [math]m =\rho_f \, V_{disp}[/math]

 

4.) You know that the volume of the cylinder is [math]V_{cyl} = H_{cyl}\, \pi \, r^2[/math]

 

5.) You know that density is equal to [math]m / V_{cyl}[/math]

 

 

Make sure that you take into account the final units that your teacher wants when you work out the answer!!!

However, if you are like myself, then you do not believe in the theory of gravity, and

The Sun and Moon heating the ocean, swelling it, becomes a much more logical answer to Tide sciences.

 

Is it that you do not believe in gravity's role in affecting bodies of water, or just gravity itself?

"Two gasoline distributors A and B are 228 miles apart on Interstate 80. A charges $0.85 per gallon and B charges $0.80 per gallon. Each charges $0.05 per mile for delivery. Where on Interstate 80 is the cost to the customer the same?"

...

The given answer is 64 miles from A.

As everyone has stated, it would seem you are missing the quantity of gas being ordered. Let's write the problem as follows:

 

Equation for A:

 

[math]0.85 \times G + 0.05 \times M[/math]

 

Equation for B:

 

[math]0.80 \times G + 0.05 \times M[/math]

 

Where [math]G[/math] is the number of gallons ordered and [math]M[/math] is the number of miles to delivery. The answer in the book is 64 miles from A. So we will cheat and use this answer which gives us the following distances:

 

Distance from A = [math]64[/math]

Distance from B = [math]228-64=164[/math]

 

Now let's substitute those values back into the equations:

 

Equation for A:

 

[math]0.85 \times G + 0.05 \times 64 = 0.85 \times G + 3.20[/math]

 

Equation for B:

 

[math]0.80 \times G + 0.05 \times 164 = 0.80 \times G + 8.20[/math]

 

How many gallons of gas should be ordered to make the answer true? We need to set both equations equal to each other and solve for G:

 

[math]0.85 \times G + 3.20 = 0.80 \times G + 8.20[/math]

 

[math]G = 100[/math]

 

If we ordered 100 gallons of gas then the cost to the us would be the same from both A and B if we were located 64 miles from A and 164 miles from B:

 

Cost from A:

 

[math]0.85 \times 100 + 0.05 \times 64 = 88.20[/math]

 

Cost from B:

 

[math]0.80\times 100 + 0.05 \times 164 = 88.20[/math]

 

 

It would cost $88.20