I found a math breakthough .I think

  • Thread starter Raza
  • Start date
In summary, the conversation discusses various math breakthroughs that the participants have made in the past, such as finding an easier way to multiply by 11 and a method for quickly squaring numbers ending in 5. They also mention some other useful tricks, such as determining if a number is divisible by 11 and a formula for finding the area of two triangles with equal sides and an included angle of 270 degrees. The conversation also includes a brief discussion about using latex code to write mathematical formulas.
  • #1
Raza
203
0
I found a math breakthough...I think

I just found another way to square a number.

Ex1: 12^2=144
but then this also equal 144
12/2=6 'Number divided by 2
12*2=24 'Number multiplied by 2
24*6=144 <---'Both number multiplied
24/4=4 <---'Both number divided

Ex2: 15^2=225
but then this also equal 225
15/2=7.5 'Number divided by 2
15*2=30 'Number multiplied by 2
30*7.5=225 <---'Both number multiplied
30/7.5=4 <---'Both number divided

Everytime you follow this thing I made up and the divide them, you always get 4.


Please don't insult me if this is stupid because I know I am a idiot. :smile:
 
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  • #2
Replace your number with x. Work through the algebra and you will see why things work out the way they do.
 
  • #3
Hehe, I remember back in the day when I was in middle school and every other day I'd think I found a break through in math.

I remember being pretty excited at figuring out that for small two digit numbers, when you multiply by 11, all you have to do is simply:
place the first digit first
add the two digits and then place them in the second place digit
place the second digit last

And voila! you get the answer. For example, 43 x 11 =4 (3+4) 3 = 473

I also remember proving that for any two triangles:
if two sides from the first triangle = two sides from the second triangle and the included angles between those two sides add up to 270 degrees, their areas were equal. Not using trig!

Also tried to figure out other ways to find PI but not everything that I tried worked :tongue2:

Hehe, the good old days :tongue:
 
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  • #4
Raza, just in case you don't realize what hurkyl said:

[tex]( \frac{x}{2})(2x) = x^2[/tex]
 
  • #5
Where do you get formulas written like that?
And I still don't get you always get the answer 4
 
  • #6
You need to use latex code to write formulas like that. If you click on any equation written in latex, it shows the code used to type it. There should be a link to the thread in the site FAQ.

[tex] \frac{2x}{\frac{x}{2}} [/tex]

multiply top and bottom by 2:

[tex] \frac{4x}{x} = 4 [/tex]

since the x's cancel
 
  • #7
I love finding things like this as well. Such as the square of any number is the the two number around it multiplied plus 1. (x-1)(x+1) + 1 = x^2 which is because (x-y)(x+y) = x^2 - y^2

I also found my own equation for pi involving dividing by tan(180/x) or something... but I can't seem to remember it!


Nothing annoyed me more than when I realized that dx/dy(x^y) = yx^(y-1) was already found. :frown:
 
  • #8
These 'rules' are really useful when doing arithmetic by head (what's it called in english?), you know, without a calculator.

I used to suck at doing calculations in my head (actually I still do. My short-term memory just leaves at that moment), but applying all these tricks really helped me alot.

For example: using (x-y)(x+y) = x^2 - y^2.
15*17 (looks tough to me) = (16+1)(16-1)=16^2-1=255.
You'll have to learn the squares from 1^2 to 20^2 by heart. By doing so you can make lots of calculations easier.
13*17=15^2-2^2=221.
This last can be used in conjunction with a trick for squaring a number that ends with 5: Take the number in front of the last 5 and multiply it with that number+1, then glue 25 to the result.
i.e. 15*15=(1)(2) '+' 25=225
65*65=(6)(7) '+' 25 = 4225
Proof: take any number x that ends in 5. Call the number before the 5 d, so that x=10d+5
(10d+5)(10d+5)=100d^2+100d+25=100d(d+1)+25.

Some of my math teachers can multiply two random two-digit numbers effortlessly. I wonder if they are just arithmetic wonders, or use a lot of these tricks...
I never had such a great feel for numbers though.
Okay, I`m going off topic..
 
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  • #9
Galileo said:
These 'rules' are really useful when doing arithmetic by head (what's it called in english?), you know, without a calculator.

I used to suck at doing calculations in my head (actually I still do. My short-term memory just leaves at that moment), but applying all these tricks really helped me alot.

For example: using (x-y)(x+y) = x^2 - y^2.
15*17 (looks tough to me) = (16+1)(16-1)=16^2-1=255.
You'll have to learn the squares from 1^2 to 20^2 by heart. By doing so you can make lots of calculations easier.
13*17=15^2-2^2=221.
This last can be used in conjunction with a trick for squaring a number that ends with 5: Take the number in front of the last 5 and multiply it with that number+1, then glue 25 to the result.
i.e. 15*15=(1)(2) '+' 25=225
65*65=(6)(7) '+' 25 = 4225
Proof: take any number x that ends in 5. Call the number before the 5 d, so that x=10d+5
(10d+5)(10d+5)=100d^2+100d+25=100d(d+1)+25.

Some of my math teachers can multiply two random two-digit numbers effortlessly. I wonder if they are just arithmetic wonders, or use a lot of these tricks...
I never had such a great feel for numbers though.
Okay, I`m going off topic..

I can multiply two digit numbers in my head faster than your average person using a piece of paper.
 
  • #10
Physics_wiz said:
Hehe, I remember back in the day when I was in middle school and every other day I'd think I found a break through in math.

I remember being pretty excited at figuring out that for small two digit numbers, when you multiply by 11, all you have to do is simply:
place the first digit first
add the two digits and then place them in the second place digit
place the second digit last

And voila! you get the answer. For example, 43 x 11 =4 (3+4) 3 = 473
Do a web search on a fellow called Trachtenberg. BTW this works for all multiplications by 11.
 
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  • #11
Also, an integer is divisible by 11 if the difference between the sum of its odd-position digits and its even-position digits is divisible by 11. E.g. 192918 is divisible by 11 because 9+9+8-(1+2+1)=22
 
  • #12
kreil said:
You need to use latex code to write formulas like that. If you click on any equation written in latex, it shows the code used to type it. There should be a link to the thread in the site FAQ.

[tex] \frac{2x}{\frac{x}{2}} [/tex]

multiply top and bottom by 2:

[tex] \frac{4x}{x} = 4 [/tex]

since the x's cancel

I don't know why you multiply by 2

Isn't it like this?

[tex]( \frac{x}{2})(2x)[/tex]

[tex]=(\frac{x}{2})(\frac{2x}{1})[/tex]

[tex]=(\frac{x}{2})(\frac{4x}{2})[/tex]

[tex]=\frac{4x^2}{2}[/tex]

[tex]=2x^2[/tex]
 
  • #13
[tex](\frac{x}{2})(\frac{4x}{2}) = \frac{4x^2}{4}=x^2[/tex]
 
  • #14
When multiplying fractions, you multiply numerators as well as denominators.
 
  • #15
If there's one thing I've learned it's that someone has always done something or thought about something well before you. I remember being so happy when I found a relationship between Pascal's Triangle and n-dimensional objects made up of one type of convex polygon. I even did a project on it and placed in the state science fair and nobody told me that it had been done before. I would have gone to ISEF except I had incredibly bad luck and had to compete for a spot in it with an RSI girl... but enough about my past :p. It wasn't until a few months later when I found some website stuck way back in the annals of the web that touched on the subject. Moral of the story: Unless you link two VERY esoteric parts of math, chances are it has been conceived and beaten into the ground.
 
  • #16
Galileo said:
You'll have to learn the squares from 1^2 to 20^2 by heart. By doing so you can make lots of calculations easier.

If you learn to square two digit numbers in your head, you don't have to memorize any.

Any 2 digit decimal number can be converted to Roman Numerals (don't worry, this is a way of thinking about numbers, so you don't actually have to write in Roman Numerals).

36 is XXXVI - you have 3 tens, 1 five, and 1 one.
39 is IXXXX - you have 4 tens and -1 ones.
44 is XXXXIV - you have 4 tens, 1 five, and -1 ones.

The idea is that it's simpler to multiply by tens and fives and then adjust to the precise value. If you modify the Roman Numeral System you learned in school:

33 is XXXIIV - you have 3 tens, 1 five, and -2 ones.

Now you can only have 5 different values for your ones: -2, -1, 0, 1, or 2.

Using that principal, you can break any two digit number into an x + y + z format.

x will be your tens, y will be your fives (it can only be 0 or 5), and z will be your ones (-2, -1, 0, 1, or 2).

Squaring the sum of (x + y + z) is:
[tex]x^2 + 2xy + y^2 + 2xz + 2yz + z^2[/tex]

You can rearrange the second term via commutative property. The fourth and fifth term have 2z in common, so you can use the distributive property to get:

[tex]x^2 + x(2y) + y^2 + (x+y)*2*z + z^2[/tex]


Now, plug in your numbers.

The tens are squared. That's easy. You squaring a one digit number and doubling the number of zeros.

Multiply x by (2 * y). 2*y will either equal 10 (if y is 5) or 0 (if y is 0). Easy step. x is multiplied by 10 or 0. Add this to your first step.

Square y. You get either 25 (y is 5) or 0 (y is 0). Add this to the sum from your second step.

The fourth term is done exactly as it is written: 1) add x + y 2)multiply by 2 3) multiply by z (When you're doing this in your head, it's easier to multiply by 2 twice than it is to multiply by 4 once). Add this to the sum from your third step (z might be negative, so this is the only tough step in the whole procedure).

Square z. Add to the sum from your fourth step.

Example: 36 = 30 + 5 + 1, so:

30^2 is 900
10*30 is 300 giving you 1200, so far.
5^2 is 25, giving 1225, so far.
35 * 2 * 1 is 70, giving you 1295, so far.
1^2 is 1, giving you your final answer of 1296.

33 = 30 + 5 - 2, so:

30^2 is 900
30*10 is 300, giving you 1200.
5^2 gets you to 1225.
35 * 2 * -2 is -140 (or -200 + 60) giving you 1085.
(-2)^2 is 4, giving you 1089.

35 = 30 + 5 + 0
30^2 is 900
30*10 gets you to 1200.
5^2 gets you to 1225.
Both the last two steps involve multiplication by 0, so they can be deleted.
 
  • #17
BobG said:
If you learn to square two digit numbers in your head, you don't have to memorize any.
That's a cool post BobG. Thanks, I hope it will be of help.

Any 2 digit decimal number can be converted to Roman Numerals (don't worry, this is a way of thinking about numbers, so you don't actually have to write in Roman Numerals).

36 is XXXVI - you have 3 tens, 1 five, and 1 one.
39 is IXXXX - you have 4 tens and -1 ones.
44 is XXXXIV - you have 4 tens, 1 five, and -1 ones.
I thought 3 was written III instead of IIV. Nevermind though, it doesn't relate to the calculation anyway.

33 = 30 + 5 - 2, so:

30^2 is 900
30*10 is 300, giving you 1200.
5^2 gets you to 1225.
35 * 2 * -2 is -140 (or -200 + 60) giving you 1085.
(-2)^2 is 4, giving you 1089.
I guess you'll just have to memorize the steps then (and lots of practice ofcourse).
The problem I have is that when there are lots of steps involved I forget what I have done, but since these steps are all easy, the chances of not forgetting increases.
 
  • #18
Galileo said:
I thought 3 was written III instead of IIV. Nevermind though, it doesn't relate to the calculation anyway.
It is. But if you use it as normally written, squaring numbers that end in 3 or 8 are a royal pain. -2 is a lot better number.
 

1. What is the breakthrough in math that you have found?

The breakthrough in math that I have found is a new algorithm that solves complex equations in a fraction of the time that it would take using traditional methods. This can greatly improve efficiency in various fields of study.

2. How did you come up with this breakthrough?

I have been working on this problem for several years, researching and experimenting with different approaches. I was able to create this breakthrough by combining various mathematical concepts and techniques in a unique way.

3. What implications does this breakthrough have for the field of math?

This breakthrough has the potential to significantly advance the field of math, particularly in the areas of data analysis, cryptography, and computer science. It can also lead to further discoveries and developments in related fields.

4. What are the limitations of this breakthrough?

While this breakthrough is a significant advancement, it may not be able to solve all types of equations. There may also be certain scenarios where traditional methods may still be more effective. Further research and testing will be needed to fully understand its limitations.

5. How do you plan to share your breakthrough with the scientific community?

I plan to publish my findings in a reputable scientific journal and present my research at conferences and seminars. I also hope to collaborate with other researchers who are interested in further exploring this breakthrough and its applications.

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