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#76 Euler Avenue » More real numbers that natural numbers? » 2010-01-16 05:34:35
- Fruityloop
- Replies: 13
I don't know how many people have seen Cantor's proof that the set of real numbers is larger than the set of natural numbers (1,2,3,4...) but he shows that you can't produce a 1-to-1 pairing of the real numbers and natural numbers like this....
You now create a new real number that is not on the list by selecting a number that differs from the first number in the first position after the decimal point and differs from the second number in the second position after the decimal point and so on. For example, the number .2738..... won't be on the list because 2 differs from 8, 7 differs from 6, 3 differs from 9, 8 differs from 2 and so on. This seems all logical and reasonable , but what if I use his same argument to show that the set of natural numbers can't be paired with itself.
If I select the new natural number 5832... I have a natural number that isn't on the list because 5 is different from 1, 8 is different from nothing, 3 is different from nothing and so on. So let's say I choose to pair the natural numbers with the set of real numbers like this...
There now seems to be a 1-to-1 correspondence between the natural numbers and the real numbers. Once you get past .9 the real numbers are simply the natural numbers reversed with a decimal point. If you try and create a new real number that isn't on the list you would be following the above logic where you were trying to show that the natural numbers can't be paired with themselves.
#77 Re: Puzzles and Games » Billiard Tables » 2010-01-16 04:33:43
Yeah, you are correct Bobbym..
I played in Vegas in 2003 and I was barred from 3 casinos in 2 days. Funny thing is, I played in Vegas in 2008 and I wasn't even barred. There is a local casino where I live that doesn't care and I can get away with betting ratios of 30-1. The only bad thing is they have 8-deck shoes and with Basic Strategy the player is at a disadvantage of about .77% It's a pretty bad game and you sometimes have to sit there for a very long time counting cards waiting for the count to go up. What never fails to amaze me is that 99.9% of the people who play backjack don't even play basic strategy, they happily throw their money away. Even the dealers are ignorant, you hit A7 and they're incredulous. Anyways enough for now...
#78 Re: Help Me ! » Chances of getting tail on a coin before 1 on a die? » 2010-01-16 03:51:02
It seems as though your problem could have two different answers depending upon whether you flip the coin first or roll the die first...
If you roll the die and then flip the coin then JaneFairfax has the correct answer..
If you flip the coin and then roll the die the answer is different..
#79 Re: Puzzles and Games » Billiard Tables » 2010-01-09 18:33:58
JaneFairfax,
Absolutely amazing! Your scores and times verge on being unbelievable! Hmmm... is there a connection between number theory and being skilled at billards?
#80 Re: Help Me ! » Tough calculus problem.. » 2010-01-07 01:23:39
OK! I finally got the answer! 
The key is to make a little rectangle underneath the ellipse, where the upper right corner touches the ellipse at the point of tangency.
In this case, I called the part from the rectangle up to where the tangent intercepts the y-axis, h.
I called the part from the rectangle to where the tangent intercepts the x-axis, b.
The slope of the tangent to the ellipse is
Both the h divided by the x-axis part of the rectangle and the y-axis part of the rectangle divided by b are equal to the slope of the tangent (the absolute value).
So we have
and we have
So the square root of the following gives the distance...
which simplifies to..
Now the length of the tangent between the coordinate axes is the square root of this, but if we take the derivative of this instead it should still give us the right answer.
So taking the derivative and setting it equal to zero and solving for x (a bit of work here)
we end up with and as the point on the ellipse where the minimum tangent touches it.
We already know the slope so we can get the equation of the line and find out the length by the pythagorean theorem.
The x-axis distance is and the y-axis distance is 6.
So squaring both and adding and taking the square root gives 9.
#81 Re: Help Me ! » Tough calculus problem.. » 2010-01-04 08:44:27
The 9 units is the length from coordinate axis to coordinate axis I believe.
This is apparently the minimum length for all of the possible tangents to the ellipse.
I don't know how to prove this.
I can get the slope of the tangent to the ellipse but then I'm stuck.
#82 Help Me ! » Tough calculus problem.. » 2010-01-03 18:10:32
- Fruityloop
- Replies: 7
I'm stuck on the following Calculus problem..
A tangent is drawn to the ellipse
so that the part intercepted by the coordinate axes is a minimum.
Show that the length is 9 units.
#83 Re: Help Me ! » i need help! » 2010-01-02 19:33:02
Try thinking about it like this...
If you have 264/5 then you have 264 fifths correct?
OK. So every 5 of those 264 fifths you can make a whole number correct?
This should lead you on the right path...
#84 Re: Help Me ! » i need help! » 2010-01-02 19:02:27
Don't feel bad. The only way to stay good at math is to practice doing problems. If you're aren't doing it, then you will have a tendency to forget. I think it's that way with pretty much everyone.
With whole numbers you multiply the denominator by the whole number part and add it to the numerator.
So 3 2/3 = 11/3
Fruityloop
#85 Re: Help Me ! » i need help! » 2010-01-02 18:50:33
You need to find the lowest common multiple of 5 and 4 (called the LCM), and then convert the fractions into that number in the denominator so you can add them.
Fruityloop
#86 Re: Help Me ! » Small question.. » 2009-10-26 16:25:07
I've been looking for some proof of this and I think I've found it..
Euclid's Elements Book III proposition 21 states..
"In a circle, the angles in the same segment equal one another."
So in Soroban's diagram, if we draw lines connecting A to C, C to B, B to H, and H to A
we immediately see that
We consequently end up with similar triangles and the segments of each line when multiplied together end up being equal to each other.
So the expression is true for ANY four points on the circumference on a circle. It seems this has been known for only about 2300 years!
Well, live and learn.
Fruityloop
#87 Re: Help Me ! » Small question.. » 2009-10-23 11:16:15
Thank you for your time and effort Bobbym. So it seems that it's just a happy coincidence of this particular problem. I thought there was some great theorem that I wasn't aware of that allowed this to be true. Pretty funny. After going through your analysis I see how to solve this. Thank you.
Fruityloop
#88 Help Me ! » Small question.. » 2009-10-22 16:37:29
- Fruityloop
- Replies: 8
This is from a problem book...
AB and CD are perpendicular diameters of a circle and are 10 units in length.
Chord CH cuts AB at K and is 8 units long.
Let x = KB and y = CK.
The book says
#89 Re: Help Me ! » Need some help finding the side of a triangle... » 2009-10-22 15:56:14
Hi Ral,
Try this...
You can easily find
Once you know those two angles (actually you only need ) you can find by the law of sines.
You will then be able to determine.
OK. So on , you now know two sides and the angle inbetween so the other side can be found by the law of cosines.
#90 Re: Help Me ! » formula for the % of increase between x & y? » 2009-10-20 11:27:36
Hi Crux,
I'm not totally sure where your confusion is but it seems as though you're taking the reciprocal of the percentage change and trying to make sense of it. Always remember that to find the percentage change take the difference between the two amounts and divide by the original amount.
6/2 = 3 so we need the recriprocal for the percentage change (1/3) = 0.3333.....
So we have 6.5/1.5 = 4.3333.....
now to find the percentage change we need the reciprocal (1/4.3333....) = 23.08%
I hope this helps,
Fruityloop
#91 Re: Help Me ! » DESPERATELY URGENT HELP NEEDED FOR THIS SOLUTION. answer in half hr. » 2009-10-14 22:44:13
Thanks for the compliment Bobbym.
One thing I find troubling about this problem is that the problem suggests that it's
impossible to rest a ladder against the upper-outer edge of a box and drag the ladder away from
the box along the ground while keeping contact with the box and wall.
It's mathematically impossible! 
I guess it's true but it somehow doesn't 'seem' right.
#92 Re: Help Me ! » DESPERATELY URGENT HELP NEEDED FOR THIS SOLUTION. answer in half hr. » 2009-10-14 22:20:03
Instinctively the problem appears to have more than one solution, but this what I have...
We will divide the ladder into two parts the upper part is (4-x) and the lower part is x.
The distance from the top of the box to the top of the ladder is y.
We have similar triangles created so...
Substituting this for y in the equation above we get..
Which gives four possible solutions for x, but the only two that work are...
So there are two possible distances up the wall the ladder can reach...
Which is the same as Bobbym's answer.
#93 Dark Discussions at Cafe Infinity » Roswell UFO crash, 1947 » 2009-10-02 18:27:08
- Fruityloop
- Replies: 0
Hi all,
Here's a website I found a few years ago that I would like to share...
http://www.roswellproof.com/
It is very hard to read, but it seems as though a few words can be made out.
VI_ _ IMS OF THE
Now it seems as though VICTIMS is the only word that fits, in which case the memo
isn't referencing a weather ballon.
Anyways, people can draw their own conclusions after viewing the website.
#94 Re: Help Me ! » Answer different from book » 2009-09-27 07:47:49
Very good! Thank you. I missed that.
#95 Help Me ! » Answer different from book » 2009-09-26 13:03:15
- Fruityloop
- Replies: 2
Hi all,
I've been working out of a calculus book doing implicit differentiation.
Find
The equation is
The book gives
But I get
I have no idea how the answer in the book was arrived at.
Anybody care to try it?
#96 Dark Discussions at Cafe Infinity » Cool cash scratch off game » 2009-09-23 09:25:04
- Fruityloop
- Replies: 0
This was from a couple of years ago.
http://www.manchestereveningnews.co.uk/ … _confusion
In the comments...
While -8 is a lower number than -7 because it is at a greater distance from zero, -7 is a *smaller* number than -8, since smaller refers to the magnitude of the number rather than its distance from zero.
So -7+20=13 and -8+20=12, so 13 is *smaller* than 12?
#97 Re: Help Me ! » probability of four from five cards being same suit » 2009-09-22 02:03:58
Hi sscrabble,
I believe you need to use the hypergeometric distribution formula. You are looking for 4 cards out of a group of 13 and 1 card from the other 39 so we have...
So the probability is 1 in 23.3.
If the odds of a flush are 508 to 1 against, the probability is 1 in 509.
So 509/23.3 gives 21.8 as the number you are looking for.
#98 Re: Exercises » Some sums » 2009-09-15 13:58:40
This was the most difficult of the three.
It is interesting that a complicated expression can be summed so simply.
#99 Re: Exercises » Some sums » 2009-09-14 08:43:25
This one took awhile to solve.
#100 Re: Exercises » Some sums » 2009-09-11 07:03:15
Math can be so frustrating!