Number question

[A Nonny Moose 8,261]

ROFL! Leave it to Dilbert.

It is easier for us, because we often have loonies and twonies wearing holes in our pockets. The weight reminds us that we should spend them for small purchases.

I have absolutely no sympathy for people who cannot make change no matter what the circumstances. I learned this in grade two. In any society that uses currency, this is a life skill.

Being innumerate is very nearly as serious as being illiterate. And with talking books and video, we may be heading that way. I worry about the little darlings, and wonder if primary school is too advanced, and should back up a little to make sure the necessary basics are in there good and solid. All the rest can be left to secondary and later.

Maybe the method of teaching multiplication should use the Tractenburg system. Using this, you can multiply two ten digit numbers together by inspection and a piece of paper. I don't recall whether there is a similar division system.

Multiplication tables should be taught and drilled up to 15. Any kid who graduates from grade six should be able to make change for any amount up to $100. And when they learn programming, they should learn how to program a change machine. It is a good exercise in the use of the modulo operator.

[Muck308 55]

When I bring change into the store with me, I count out three quarters, two dimes, a nickel, and four pennies. Why this combination? Well, with those coins, you can assemble any cent amount from .00 to .99, which means I can pay exactly and get no change, or at least pay the cents part accurately and get only bills in change.

I've never met anyone else who does that! I thought I was the only one. Although I don't carry that combination of coins on my person very much any more, but when I was younger I could always be found with said coins on me.

As far as confusing cashiers goes, I often find myself with, let's say a $20 bill. I want to buy gas and as much as possible I might add but I also want to buy, let's say a Pepsi. So I ask them to ring me up the Pepsi and when it comes to $1.59 I say

"$18.41 on pump whatever as well please."

I often get confused looks or even asked. "How did you do that?"

Because I'm not an idiot, that's why. Of course that's never my response. That would be far too rude.

[Easy Bakes 253]

Win. That's an engineer's logic alright.

For anyone who doesn't get it:

For anyone who doesn't get it:

$7.14 - $1.89 = $5.25, change which can be given with just one bill and one coin. If Dilbert were the cashier, he'd consider that making his job easy.

However if old dilbert had $7.14 for a $1.89 order he would have given him a $5.00 2 $1.00 a dime ad 4 pennies correct? So why give him the 5? he would just give it back to you. give him 2 ones and 14 cents and get a single quater back.

mabey he had 7 ones. and wanted a 5 back...........but would have been

screwed at the vending machine at work.

  Edited April 23, 2011 by Easy Bakes  

[Duke87 1,070]

Multiplication tables should be taught and drilled up to 15.

We were taught and drilled up to 12. But this is the sort of skill that can be very "use it or lose it", and with calculators, people don't use it.

As far as confusing cashiers goes, I often find myself with, let's say a $20 bill. I want to buy gas and as much as possible I might add but I also want to buy, let's say a Pepsi. So I ask them to ring me up the Pepsi and when it comes to $1.59 I say

"$18.41 on pump whatever as well please."

I often get confused looks or even asked. "How did you do that?"

Sensible, I suppose, although I would never do such a thing for two reasons:

1) I do not typically pay for gas with cash, it's debit at the pump unless that doesn't work (sometimes they make you go inside).

2) I always fill my tank whenever I buy gas, because if I don't it screws up my fuel economy calculations.

[A Nonny Moose 8,261]

Another trick of numeration for people who can remember their times table. To square any two digit number ending in five, take the tens digit and multiply it by itself + 1, put that number down with 25 behind it.

E.g.

95 x 95 = 9025

15 x 15 = 225

(10x + 5)² = 100x(x+1) + 25

Try expanding the LHS algebraically and you get the RHS.

[blade2k5 1,523]

Of course that's never my response. That would be far too rude.

I'm not. I tell them exactly what I think, like, learn how to add/subtract, it's not difficult. If they find that rude, it's their problem, not mine.

[A Nonny Moose 8,261]

Of course that's never my response. That would be far too rude.

I'm not. I tell them exactly what I think, like, learn how to add/subtract, it's not difficult. If they find that rude, it's their problem, not mine.

I agree with this. If nobody tells them how deficient they are, how will they ever know? Sometimes we are just too damned Politically Correct or too polite.

[blade2k5 1,523]

I agree with this. If nobody tells them how deficient they are, how will they ever know? Sometimes we are just too damned Politically Correct or too polite.

Exactly. PC is for gutless weak-minded fools.

[Meg 368]

I agree with this. If nobody tells them how deficient they are, how will they ever know? Sometimes we are just too damned Politically Correct or too polite.

I agree that political correctness can be over done but I haven't seen "too polite" in years. People seem to be getting ruder all the time.

[Duke87 1,070]

Being rude accomplishes nothing. An idiot isn't going to stop being an idiot no matter how many people call him one.

Besides, you can explain to someone what you're doing without being rude about it. It only reflects negatively on your own intelligence if you have to resort to insults to get a point across.

[astronelson 339]

One often finds it pays to be nice to the waiter, although one does hope that one is polite because of how one is raised, not for any gain. Impoliteness and rudeness may get you places, but it is not a means of transport I endorse, and I avoid it whenever possible. One finds the company on the journey and the destination quite distasteful.

With respect to a previous comment, while I do agree that political correctness can be taken too far (although I have not seen it myself outside parodies), I disagree that one can be too polite.

May we now please return to matters of math? These problems of politeness really belong in another thread.

To start us off again: Take a circle with diameter 1. Obviously the circumference is pi. Now draw a square around it. Obviously its perimeter is 4.

Cut a square out of each external corner such that the interior angle is on the circumference of the circle. The perimeter is still, obviously, 4.

Repeat this last step ad infinitum. The limit as one approaches infinity of this curve is the circle. However, the perimeter is 4 and the circumference pi. Can anyone resolve this apparent paradox?

[A Nonny Moose 8,261]

To start us off again: Take a circle with diameter 1. Obviously the circumference is pi. Now draw a square around it. Obviously its perimeter is 4.

Cut a square out of each external corner such that the interior angle is on the circumference of the circle. The perimeter is still, obviously, 4.

Repeat this last step ad infinitum. The limit as one approaches infinity of this curve is the circle. However, the perimeter is 4 and the circumference pi. Can anyone resolve this apparent paradox?

The fallacy is in the word obviously. If you produce a limits statement on your reducing figure, you will discover the limit is pi.

If you want a pair of docs, I'd suggest you look in the nearest harbor.

Now, here is one for all you geometers out there. When can a circle be transformed into a straight line?

  Edited April 24, 2011 by A Nonny Moose  

[astronelson 339]

"Obviously" was only in it for the parts that were true.

I was not sure of the solution myself, but as infinity was involved, something was, most certainly, up.

To be honest, I searched the internet for "Troll Math" and chose something that gave an interesting result without division by zero.

If you want a pair of docs, I'd suggest you look in the nearest harbor.

I prefer a PhD and an MD on opposite sides of a benzene ring. A JD would suffice, although that is not quite as well-known.

With all three, one can have simultaneous ortho-docs, para-docs, and meta-docs. A paper on the last would be interesting, although one wonders what to do with the abstract.

[A Nonny Moose 8,261]

"Obviously" was only in it for the parts that were true.

I was not sure of the solution myself, but as infinity was involved, something was, most certainly, up.

To be honest, I searched the internet for "Troll Math" and chose something that gave an interesting result without division by zero.

If you want a pair of docs, I'd suggest you look in the nearest harbor.

I prefer a PhD and an MD on opposite sides of a benzene ring. A JD would suffice, although that is not quite as well-known.

With all three, one can have simultaneous ortho-docs, para-docs, and meta-docs. A paper on the last would be interesting, although one wonders what to do with the abstract.

Well, I understand that a JD can be on both sides at the same time. Shakespeare had a place for equivocators, eh? (The Scottish Play, the gatekeeper's scene). However three Greek prefixes in the same benzene ring could really compound the problem. Could they be accused of glue sniffing? Maybe they could pair up with the weird sisters (hand in hand, posters of the sea and land).

Afterthought: Could Schroedinger's cat have a JD?

  Edited April 24, 2011 by A Nonny Moose  

[Jeffmaster223 24]

My! so many equations! thank you, people!

And yes, i could have looked it up, but what's the point

of it? it's not too fun! or... enjoyable!

-Anim.

[astronelson 339]

Well, now this thread has been revived, there is a question I would like to ask, open to anyone who wishes to try it:

Can you prove that for any continuous periodic function f defined on the real numbers, you can draw a horizontal line (parallel to the x-axis, that is), that intersects the function at two points separated by some arbitrary distance k, where k is a real number? That is, can you prove that for any real k, there exists real x such that f(x)=f(x+k)?

I have a vague idea of a proof, but do not know how to express it in words.

[Meg 368]

We were taught and drilled up to 12. But this is the sort of skill that can be very "use it or lose it", and with calculators, people don't use it.

Can you prove that for any continuous periodic function f defined on the real numbers, you can draw a horizontal line (parallel to the x-axis, that is), that intersects the function at two points separated by some arbitrary distance k, where k is a real number? That is, can you prove that for any real k, there exists real x such that f(x)=f(x+k)?

Speaking of "using or losing" a skill, I am several decades past knowing how to do that.

[hamsterTK 93]

God forbid if the power goes out and they can't use the register.

The register isn't just a big calculator, it does inventory too now. Modern stores have thousands of items and high volumes of customers, and the computers reduce the amount of time and labor needed to keep it all organized while preventing human error and theft.

I have a feeling the store will lose less money by pausing for 5 minutes during a network downtime than 5 hours of overnight stocker overtime after doing it the old fashioned way.

Other reasons would be that most people pay with plastic now and also its the employee punchclock.

  Edited July 5, 2011 by hamsterTK  

[ROFLyoshi 1,760]

I have worked in a pizza shop for the last few years and it surprises me how many of my coworkers needed to use a calculator or the till to work out change. I do it mentally quickly as follows.

Order is $15.80, you're given a $50.

Take out a 20 (50 - 20 = 30)

Take out a 10 (30 - 10 = 20)

Take out two 2s (20 - 4 = 16)

Take out a 20c (16 - .2 = 15.80)

It looks complicated, but its easy.

Also if I order something, I will try to give them the change so I will only get dollars back. (I'll pay for something that cost 9.30 with $20, 20c and a 10c). I swear I've confused cashiers by doing that.

[A Nonny Moose 8,261]

True, and more than true. I do the same thing, but I have the courtesy to explain what I did and what I expect.

When I was teaching C, I used to use a class problem to create a change machine (virtual) to make change for any amount up to $5. (Canada has coins to make that work.)

It is an exercise in using the modulo operator and the integer divide operator. You would be surprised how many couldn't do it without a hint. A good programmer can make this happen using a recursive function.

When I was in about grade 3, I think, we learned how to make change. I guess that's not taught any more.

  Edited July 6, 2011 by A Nonny Moose  

[bigro 165]

*looks at horribly horrible and ugly equations*

one day I'll know I am only in year 11 after all. Now for the ultimate reply to nigh on every subject brought up here. (cept the maths ones...cos y'know, I can't)

First , calculators are super, I got my first one in grade 6. Second, I learnt all that silly 'do it in your head' stuff in junior school and to an extent middle school. Senior however I have yet to be asked to write out a long division equation or look up sin.theta. Third, cashiers should be able to do all the maths and stuff in thier heads. (barring something like a cashier at a furniture store where they are actually working with numbers over 100) and it makes me feel all special that I am better at shorthand maths than an average Aerican and I spent reception through year 10 in Alice Springs (and thats saying something, pick your crap up America) And last but not least, cashiers in my current area are smart no fun at all. No really, they are. For some reason this area of Adelaide (just outside actually) is just where all the nerds congregate...now I feel real smat. (misspelt on purpose)

also, not done bieng confused by some of these equations, might come back later after I learn some.

[hamsterTK 93]

Honestly I think its just brain fart when you have to count out change to someone in public. Today I had to buy something with change and gave the guy too many dimes. I believe I was distracted thinking of something else and somehow adding 10 cents to 1.25 resulted in the mental equivalent of a blue screen

Doesn't mean I can't do basic addition. And I can do long division on paper.

Its kind of like being forced to read something aloud, which feels like walking and chewing gum at the same time when you are used to reading very quickly to yourself by skimming. I'm sure its obvious that I am literate, otherwise I wouldn't be typing coherent posts on this forum.

  Edited July 6, 2011 by hamsterTK  

[Duke87 1,070]

Can you prove that for any continuous periodic function f defined on the real numbers, you can draw a horizontal line (parallel to the x-axis, that is), that intersects the function at two points separated by some arbitrary distance k, where k is a real number? That is, can you prove that for any real k, there exists real x such that f(x)=f(x+k)?

By definition, if a function is periodic, then it can be said to consist of an infinite number of identically shaped segments laid end to end parallel to the x axis. If the segments are all identical, then logically each ordinate (y value) the curve achieves in one segment must be achieved in every other segment. If there are an infinite number of segments, then logically each of these ordinates must be achieved an infinite number of times. And if any ordinate is achieved an infinite number of times, then it must therefore be possible to draw a horizontal line at that ordinate which intersects the curve not only twice, but an infinite number of times.

Or, we can come at it the other way around:

Let's assume one can draw a horizontal line which only intersects the repeating curve once. If it only intersects the curve once, then the ordinate at which the line is drawn must only be achieved once. Therefore, we have a repeating function which achieves some ordinate for only one value of x, Q.E.A. Thus, any horizontal line which intersects the function must intersect it at least twice.

This is of course just using logic, otherwise known as common sense. If you want a proof that actually uses math... I can't help ya.

  Edited July 6, 2011 by Duke87  

[astronelson 339]

I'm not asking for a proof that it intersects the line twice - I'm asking for a proof that no matter what arbitrary distance k you choose, there exist two points on the curve separated by that arbitrary distance have the same value. For example, prove that there exists some x for which Sin[x]=Sin[x+1], but generalize for all periodic functions and all arbitrary distances k.

I specified periodic continuous functions, since one can easily come up with counterexamples if either condition is not met (x is continuous but not periodic, and tan[x] is periodic but not continuous, and it doesn't hold for either, but I can't think of a counterexample for any periodic continuous functions).

[Shadow Assassin 336]

I haven't looked at a trigonometry book lately but do they still come with those tables in the back? or are students expected to use their calculators instead of looking things up?

My old high school maths text book (printed in 2001 with new editions every year, which we had to buy, coming out til 2005) had everything we needed in the back. Trig values, etc.

Our scientific calculators were invaluable though - nowhere near as advanced as a graphics calculator, but they did the job for what we needed to do. One of the most important skills I had to learn (with regards to trig) was how to convert from pi to radians in my head. With the help of the aforementioned trig values in the back of the book, this was easy to do. The scientific calculator (which I haven't used in about five years actually XD) was helpful in ensuring errors were removed from the value... after all, I (and most people) could only do an approximation unless I was a mathematical genius or something

I looked at a new maths text book for some people who were just starting high school... nothing like that. I feel sorry for them.

[A Nonny Moose 8,261]

Astro, your problem lies in the domain of set theory, and it has been so long since I did that, I won't even try to dredge it up. The logical proof by the Duke should be sufficient, but I suppose one could take the second conjecture and, by induction, cough up a suitable set of expressions. I used to like inductive proofs, but they make my brain hurt.

The educaterers will tell you that, with calculators generally available, all that memory work in basic arithmetic is no longer needed in the curriculum. Let me give you an example of why you should make sure your kids can do some mental arithmetic.

One of the more glamorous sports is sailing around the world single handed. So you set off, and get down to Cape Horn to round it when you are pitch-poled in a storm and all your batteries and hand-held gear are destroyed by sea-water. You have left a set of Navigation tables, Charts, a Pilot Chart for the southern ocean, a sextant and a mechanical alarm clock with only an hour hand left. But your arithmetic is poor. You also have a copy of Nathaniel Bowditch's Compendium of Navigation. You have ample supplies, but no radio or other electronics. Now what?

The second volume of Bowditch has a summary of arithmetic and mathematics of navigation written for the semi-literate deck hand, and a set of artimetic and mathematical tables containing trigonometric functions including their base 10 logarithms. End of hint.

If you want to read the exciting conclusion it will be found in "Sailing Alone Around the World" by Joshua Slocum (1906).

[Jeffmaster223 24]

that, astronelson, is beyond my mathematical knowledge.

it's still a good question.

[A Nonny Moose 8,261]

Are there any uninteresting numbers?

[Easy Bakes 253]

Are there any uninteresting numbers?

iv always disliked 73.

[Alejandro24 575]

Are there any uninteresting numbers?

Maybe the Interesting number paradox have the answer to that question.

It can be explained with basic Logics -I've learned this in the class of Logics and Sets, with the Morgan Laws-:

Supossed there are two sets: Interesting Numbers and Boring Numbers. In the set Interesting Numbers there are many numbers because they have a property -mathematical or cultural-, 1, 2, 3, ... 100, 200, 666, 1000, pi, fi, etc... At the other side, there is the set Boring Numbers, with the numbers without any special property. In the set Boring Numbers there is a number that is the most little or minor. That number is moved to the Interesting Numbers set only because have the property of being the minor of the Boring Numbers set. But again there is another minor number in the Boring Numbers set, that is moved too. This action repeats untill there is one number in the Boring Numbers set, that have the special property of being just the only one boring number. This number is moved to the Interesting Numbers set, leaving the Boring Numbers set empty.

What means this? Simple, this demonstration concluded in that the first affirmation -Interesting Numbers and Boring Numbers sets- is false. Then the conlusion is: There are not any uninteresting number.

But there is a detail: the property Interesting is not a formal mathematical property. So, the demostration is just prooved with reductio ad absurdum. We can't divide a mathematical set just with the property of being interesting or uninteresting -boring-, that is not the same to use a mathematical property as a even number, odd number or prime number.

  Edited July 23, 2011 by Alejandro24