Galactic Matyhematics.

@George,

I've always thought of 17 as slightly demonic.

@engineer,

Given your avatar, I shall now regard 17 in a new light.

0 Replies

@engineer,

Quote:

When you "cast out" the 17's, you are dividing by 17 and finding the remainder except that you are doing it in a long and painful way. It would take you far longer to repeatedly subtract 17 from 100 than to just do the problem normally. And you have to repeat this 17 times? Sorry, that is just ridiculous. I can see why you were loathe to show it to us.

well, you haven't really worked with it, because if you do it becomes extremely fast. I had no problem showing at all!

Quote:

Very, very, very mechanical indeed.

That is a strange reply! it is extremely fast but you don't say anything about that!

0 Replies

Another example is (try this by hand it is very fast compared with the conventional way)

1/7=?

First we see that the recurring decimal has to end in a '7' !

the first remainder is 3, so we have a Geometrical Progression 1: 3

and we find as the remainders:

3, (3x3=9) becomes 2, 3x2=6, 3x6=18 (becomes 4). 3x4=12 (becomes 5), 15 (becomes 1) , and now we can stop because 7 x 1=7!

we write down the remainders:

3 2 6 4 5 1

multiply each with 7 and write down the right hand digit.

142857

And we have our answer:

1/7=0,142857

You really have to work with it, as I said at the beginning, otherwise , you won't get it. Just DO it.

But there is also a beautiful way to do any division in VM and do
a 'straigh division' I will come to that.

1 Reply

@Quehoniaomath,

Quehoniaomath wrote:

That is a strange reply! it is extremely fast but you don't say anything about that!

Because it isn't very fast, it is very slow and tedious. Your example of 1/7 is the perfect example. First you find a remainder, then you do SEVEN division problems, writing down the remainders, then you so SEVEN multiplication problems and select the right hand digit. That's not beautiful, that is tedious. If you do it the standard way, you only have to do seven division problems and you are done.

10/7 = 1 r3
30/7 = 4 r2
20/7 = 2 r6
60/7 = 8 r4
40/7 = 5 r5
50/7 = 7 r1

.142857

So much simpler than what you posted and it can be used for every division problem without remembering that "the recurring decimal has to end in '7'!"

2 Replies

@engineer,

How many weeks will people on this thread go on trying to talk to a wall? No matter how well people have demonstrated and proven the method is flawed, this guy is denial. He won't own up.

2 Replies

@engineer,

No, it is very fast. AND you don't get it at all! I didn't do division, that is the whole point!

BUT,

the real reason it seems slow to you is becaue your indoctrination with the 'conventional math'.
I had the same problem and ONLY after working a lot with VM, you start to lose the indoctrination and see that VM is much faster, but you really have to UNLEARN the conventional way. And that takes time and experience, there is no other way!
Children who weren't indoctrinated with the conventional crap 'grasp' VM iimmediately,
You are just a bit biased. Wink

So, start using it to understand it. There is no other way,
It really takes time to write things down, but once you can do it, you can even do it very fast , in your mind alone! But you need three things: experience, experience, ezxperience. Wink

You can discuss all you want, you can only understand it if you DO it.
just like riding a car

And, as I said, it is alo used in the field of IT! if there isn't something to it, why do they use it then?

exactly! there is something to it.

1 Reply

@Ragman,

Quote:

How many weeks will people on this thread go on trying to talk to a wall? No matter how well people have demonstrated and proven the method is flawed, this guy is denial. He won't own up.

So, you never used it as well? Give it a try.

0 Replies

Now, for something different, calculate

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5) the conventional way!

Come on , give ma laugh! Wink

have fun!

1 Reply

@Ragman,

Quote:

How many weeks will people on this thread go on trying to talk to a wall? No matter how well people have demonstrated and proven the method is flawed, this guy is denial. He won't own up.

And, it is the other way around of course, but you won't get it. Wink

0 Replies

@Quehoniaomath,

Quehoniaomath wrote:

No, it is very fast. AND you don't get it at all! I didn't do division, that is the whole point!

Quehoniaomath also wrote:

and we find as the remainders:

3, (3x3=9) becomes 2, 3x2=6, 3x6=18 (becomes 4). 3x4=12 (becomes 5), 15 (becomes 1) , and now we can stop because 7 x 1=7!

Yes, you are doing division. That is how you get remainders. If you say you compute them via continuous subtraction, then you are doing slow tedious division, but you are still doing division.

Quehoniaomath wrote:

the real reason it seems slow to you is becaue your indoctrination with the 'conventional math'.

The real reason it seems slow to me is that you are doing the exact same steps I would do and then another set of steps and finally a third set of steps. In real math, doing one step is easier than doing three steps of equal size.

1 Reply

@Quehoniaomath,

Quehoniaomath wrote:

Now, for something different, calculate

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5) the conventional way!

Let's change the last denominator to (2x+11) since you've used your (x+5) example already. Unfortunately your tricks don't work if the values don't center around the same value so that will be beyond you. Too bad the real world doesn't rotate around fixed values.

1 Reply

@engineer,

Don't chicken out now, mate!

Quote:

Now, for something different, calculate

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5) the conventional way!

1 Reply

@engineer,

no, I don't do division. you only THINK I do.

You ar really blinded by the conventional method.

0 Replies

And, as I said, it is used in the field of it, so there really is something to it!

1 Reply

the way you are reacting is this, I write it again:

Quote:

There is a world-wide debate currently raging about the efficacy of Vedic Mathematics versus the crumbling foundations of Western Mathematics.
Generally speaking, the theorems we all learned at school are not wrong but clumsy. Some of the Western geometrical formulae are certainly wrong or inadequate: for example, the formulae for sphere packing in the higher dimensions increase up to the 6th Dimension then suddenly decrease for higher dimensions, which is simply absurd.
Unfortunately, some die-hard senior mathematicians in an attempt to protect the crumbling foundations that they now stand on feel threatened by the lightning quick mental calculations of the Vedic seers, and go to great lengths to deride Vedic maths as a "bag of tricks". And of course, many insecure teachers worldwide are afraid to rewrite all their course material.

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@Quehoniaomath,

Quehoniaomath wrote:

no, I don't do division. you only THINK I do.

Fine, then I didn't do division either and still got the answer in a third of the time.

Quehoniaomath wrote:

Don't chicken out now, mate!

Quote:
Now, for something different, calculate

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5) the conventional way!

-3.5. The standard way is to look at how you arranged the values on either side of a constant, laugh and provide the answer. No math of any sort required other than noticing that you used 3&4 on one side and 2&5 on the other. But the problem is

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(2x+11) using VM.

You've once again posted words instead of "VM". Can you not solve this very simple change?

3 Replies

@Quehoniaomath,

Quehoniaomath wrote:

And, as I said, it is used in the field of it, so there really is something to it!

You've said lots of things but since they fail to stand up to inspection, I don't know why you would expect us to believe something on your word. What "field of it" do you want us to believe is so restrictive that these rules would have application?

1 Reply

@engineer,

Quote:

Now, for something different, calculate

1/(x+4) + 1/(x+3)=1/(x+2) + 1/(x+5) the conventional way!

and please write it all out!

Thank you,

0 Replies

@engineer,

Quote:

You've once again posted words instead of "VM". Can you not solve this very simple change?

what do you mean with this, I don't understand.

1 Reply

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