# Think you know math?

Lissa
Lissa
Explain why this works...

Y = X

multiply both sides by X:

YX = X^2

subtract Y^2 from both sides:

YX - Y^2 = X^2 - Y^2

some factoring:

Y (X - Y) = (X + Y) (X - Y)

devide both sides by (X - Y):

Y = X + Y

subtract Y from both sides:

0 = X

thus, Y = 0.

Also, solve the following:

You have a metal bar that is exactly 1 mile in length (5280 feet or 63360 inches). The bar is pinned at both ends so it cannot expand along the lenght the bar. Now, the bar expands by 1 inch. How high does the bar bow?

Now, no looking on other people's paper, this is test after all...it's open book, open notes, closed friends...

Nystul
Nystul
Explain why this works...

Not sure why it "works" but I can tell you why you shouldn't count on Y = 0, X = 0, for all Y=X

(snip to Lissa's trick...)

devide both sides by (X - Y):

WOAH!!! First of all, you misspelled divide But more importantly, since Y = X, X-Y = 0. You wouldn't really want me to divide something by zero now would you. Pretty sly though; I read through the problem several times before I noticed this.

As for the second question, I think I left the solution to that one in my math books 2 or 3 years ago. But if it pops into my head I'll post another reply

-N

Joined: August 29th, 2001, 4:26 am
Explain why this works...

Y = X

multiply both sides by X:

YX = X^2

subtract Y^2 from both sides:

YX - Y^2 = X^2 - Y^2

some factoring:

Y (X - Y) = (X + Y) (X - Y)

devide both sides by (X - Y):

Y = X + Y

subtract Y from both sides:

0 = X

thus, Y = 0.

Also, solve the following:

You have a metal bar that is exactly 1 mile in length (5280 feet or 63360 inches). The bar is pinned at both ends so it cannot expand along the lenght the bar. Now, the bar expands by 1 inch. How high does the bar bow?

Now, no looking on other people's paper, this is test after all...it's open book, open notes, closed friends...
The divide by zero is the oldest trick in the book. And one that can be used to "prove" that 1=0 or 2=0 or x=5123 if used properly.

As for the length of Steel question. I would have no idea. However I did come third in the state in a maths competition last year (beating a HEAP of grade 12 students) that had impossibly difficult questions on it (I got 3.5 out of 5 right on a 3 hour exam...).

Joined: September 24th, 2001, 4:04 am
I always liked using 0/0. By the rule of zero divided by anything is zero: 0/0=0. By the rule of anything divided by itself is one: 0/0=1. Thus, 0/0=0=1. 1+1=2 and 1=0, thus 0+0=2, and thus, 0=1=2, from the obvious continuation of this all numbers can by "proven" to be equal.

-AnorX

-AnorX

Joined: August 29th, 2001, 4:26 am
0/0 is actually unsolvable. try it in your calculator. If you want to think of it this way, to how many people can you give nothing and still have nothing? The answer is any amount. similarly with n/0 where n is any number. How many people can you give nothing if you have 3 things to give?

Not a very clear analagy, but who cares. It's fun though, confusing things like that .

Occhi
Occhi
Explain why this works...

Y = X

multiply both sides by X:

YX = X^2

subtract Y^2 from both sides:

YX - Y^2 = X^2 - Y^2

some factoring:

Y (X - Y) = (X + Y) (X - Y)

devide both sides by (X - Y):

Y = X + Y

subtract Y from both sides:

0 = X

thus, Y = 0.

Also, solve the following:

You have a metal bar that is exactly 1 mile in length (5280 feet or 63360 inches). The bar is pinned at both ends so it cannot expand along the lenght the bar. Now, the bar expands by 1 inch. How high does the bar bow?

Now, no looking on other people's paper, this is test after all...it's open book, open notes, closed friends...
You have a metal bar that is exactly 1 mile in length (5280 feet or 63360 inches). The bar is pinned at both ends so it cannot expand along the lenght the bar. Now, the bar expands by 1 inch. How high does the bar bow?

1 inch over 63360 inches = 1.57828_ X 10^-5

Somewhere, I am sure one of my old text books has the forumla for the length of an arc and the fraction of a circle that it represents, however, if the bar is indeed one mile long, and fixed in position, the bowing will, or would, be insignificant to any functional use of a mile long bar of metal . . . such as a rail for a train that had somehow been either welded, or otherwise fused, in defiance of all common sense and good civil engineering practices.

The difference between an engineer and a mathematician is that the engineer frequently rounds to three(or four) significant digits, and the bridge still stays up; while the mathematician tries to solve the problem of the bridge truss dimensions to an uneconomically produceable precision.

8^P

The Rube

Lissa
Lissa
0/0 is actually unsolvable. try it in your calculator. If you want to think of it this way, to how many people can you give nothing and still have nothing? The answer is any amount. similarly with n/0 where n is any number. How many people can you give nothing if you have 3 things to give?

Not a very clear analagy, but who cares. It's fun though, confusing things like that .
When you hit 2nd semester college Calc, you find out that 0/0 can be anything and CAN be solved using L'Hospital's Rule. By using L'Hospital's Rule, you can determine one of three possible answers for 0/0, those answers are 0, infinity, and a numeric value.

Only hit on the bars...think Trig...

Lissa
Lissa
You have a metal bar that is exactly 1 mile in length (5280 feet or 63360 inches). The bar is pinned at both ends so it cannot expand along the lenght the bar. Now, the bar expands by 1 inch. How high does the bar bow?

1 inch over 63360 inches = 1.57828_ X 10^-5

Somewhere, I am sure one of my old text books has the forumla for the length of an arc and the fraction of a circle that it represents, however, if the bar is indeed one mile long, and fixed in position, the bowing will, or would, be insignificant to any functional use of a mile long bar of metal . . . such as a rail for a train that had somehow been either welded, or otherwise fused, in defiance of all common sense and good civil engineering practices.

The difference between an engineer and a mathematician is that the engineer frequently rounds to three(or four) significant digits, and the bridge still stays up; while the mathematician tries to solve the problem of the bridge truss dimensions to an uneconomically produceable precision.

8^P

The Rube
Think Trig....

Charis
Charis
Well, you're both right. The assumptions we're expected to make include a weightless beam with no sag, and assume that the rod is in fact straight, and not following the curvature of the earth. But I think Occhi would be surprised to learn that given these assumptions...

That bar is going to bow not a trivial amount, but...

154 inches. Over two Occhi's in height!!

How is that?
Consider p as the new length (or perimeter), x as the original straight length, and draw in a circle of radius r, and height y from the center to the line where the unbent beam lies, and t the half angle defining this arc. (For a picture, draw an ice cream cone!)

Then...
p = 2 t r
y = r cos t
x = 2 r sin t

Take h as the bow height, r - y, and d as the change (delta) in length, p - x.

We get:
h / d = (1 - cos t) / 2 (t - sin t)

Taking the Taylor series expansion, we get:

h / d = 3 / 2t
And since x = p (sin t) / t, t = sqrt(6(1-x/p))
Thus for small t,

h = d * 3/2 * 1/sqrt(6*(1-x/p))
For x = 63360, p = 63361, we get:

h = 154.1 inches.

The mathematician of course takes more than one term in the Taylor series and gives a closed form solution which is more 'correct' but doesnt actually give you a number! I'm tempted to say the 'lazy' engineer guesses at the answer thinking it's insignificant, but in fairness, the assumptions in the solution are ones that a good engineer should scoff at.

Charis

PS Hey KoP! Could you double check this for me please??? This number seems pretty large!! If you're busy, perhaps Friar could?

PPS As for question 1, the divide by zero is, as already pointed out, the source of the error.

Lissa
Lissa
Pythagorean's Theorm works well for an estimate...

By using Pythagorean's Theorm, first estimate is 178 inches...or almost 17 feet...