# Square Through Squares

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It seems like a very complicated way of calculating squares.

And it will inly work for numbers up to 7, if I understand it correctly.

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It seems like a very complicated way of calculating squares.

And it will inly work for numbers up to 7, if I understand it correctly.

yes complicated but a new way on X-Y Axis why u saying it will work up to 7 pl explain.

square of 7

as per formula

(7-2)th(3rd squ) +(7-3)squ.(7-3)squ

24 + 25=49

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27 minutes ago, mathspassion said:
why u saying it will work up to 7 pl explain.

square of 7

as per formula

(7-2)th(3rd squ) +(7-3)squ.(7-3)squ

24 + 25=49

Well, I'm not quite sure what you are doing there (you seem to have skipped some steps). But as far as I can see, the (7-2)th square (i.e. the 5th square) is empty.

So, when you write "(7-2)th(3rd squ)" how do you get from there to "24"?

Edited by Strange
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So... you've converted 1 number lookup (the N) and 1 multiplication (N*N) into 4 number lookups (have to find the numbers on the square snake thing), 2 multiplications, and 1 addition. I guess I fail to see any practical reason for it, tho that has been said about a lot of math.

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20 hours ago, Strange said:

Well, I'm not quite sure what you are doing there (you seem to have skipped some steps). But as far as I can see, the (7-2)th square (i.e. the 5th square) is empty.

So, when you write "(7-2)th(3rd squ)" how do you get from there to "24"?

very easy 5th(at y) square to 3rd squ(at x) count the nos of points u will get 24.

9 hours ago, Bignose said:

So... you've converted 1 number lookup (the N) and 1 multiplication (N*N) into 4 number lookups (have to find the numbers on the square snake thing), 2 multiplications, and 1 addition. I guess I fail to see any practical reason for it, tho that has been said about a lot of math.

sorry no any converted.............. it is very hard to get formula ......................I took  years to get it. It is not a Manipulation dear.

Dear Bignose  ...........Very simple to write anything(there is something wrong with peoples)  but when u do something new it takes  time and lot hard work .

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38 minutes ago, mathspassion said:

very easy 5th(at y) square to 3rd squ(at x) count the nos of points u will get 24.

There is no value in the 5th(at y). Or are you not using the numbers in your grid? If not, I don't understand what you are doing.

What do you mean by "count the nos of points"?

Perhaps you need to show a complete example; i.e. all the steps not just the first and the last.

Edited by Strange
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On 5/24/2018 at 3:51 PM, Strange said:

There is no value in the 5th(at y). Or are you not using the numbers in your grid? If not, I don't understand what you are doing.

What do you mean by "count the nos of points"?

Perhaps you need to show a complete example; i.e. all the steps not just the first and the last.

sir very strange have u not seen the two examples .......................... example also quoted of 5 and 4 .

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2 hours ago, mathspassion said:

sir very strange have u not seen the two examples .......................... example also quoted of 5 and 4 .

I can't follow your examples. Can you show how you get from:  (7-2)th(3rd squ) +(7-3)squ.(7-3)squ to 24 + 25

Presumably,  (7-2)th(3rd squ) = 24 ?

But why?

What is (7-2)th and what is (3rd squ) and why?

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3 hours ago, Strange said:

I can't follow your examples. Can you show how you get from:  (7-2)th(3rd squ) +(7-3)squ.(7-3)squ to 24 + 25

Presumably,  (7-2)th(3rd squ) = 24 ?

But why?

What is (7-2)th and what is (3rd squ) and why?

very strange and very simple why do u not try to understand ... in your first quote u wrote it is very complicated it means u have read all the things with example now what u want to ask i am unable to tell u ...

now  (7-2)th means 5th square ,(3rd squ) mean third square now count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square.ok

dimag per jor lagaooo........

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4 hours ago, mathspassion said:

now  (7-2)th means 5th square ,(3rd squ) mean third square now count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square.ok

Or let me try... start from origin to 3rd square  to 7th square : OK, I start from the origin, count to the 3rd square (on Y, presumably) that is 3, then I count across to the 7the square (on X) and that is another 4. So I make that 7 total. Or, if I multiply them, 12.

Maybe I am being very slow, but I don't get it.

You are obviously counting them in a different way. Can you explain how?

Also, why do you say "count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square"? What happened to the "7-2th" square?

If you don't want to explain, I'll just have to give up trying to understand it.

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Yeah, I've used grid lookup tables for caculators before, but explanation isn't making much to me either and not seeing why not just use a single lookup instead.

4,4 -> 16

5,5 -> 25

6,6 -> 36

etc.

Can do more too. Easily add a seperate signal for addition, subtraction, multiplication.

Edited by Endy0816
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On 5/26/2018 at 9:37 PM, Strange said:

Or let me try... start from origin to 3rd square  to 7th square : OK, I start from the origin, count to the 3rd square (on Y, presumably) that is 3, then I count across to the 7the square (on X) and that is another 4. So I make that 7 total. Or, if I multiply them, 12.

Maybe I am being very slow, but I don't get it.

You are obviously counting them in a different way. Can you explain how?

Also, why do you say "count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square"? What happened to the "7-2th" square?

If you don't want to explain, I'll just have to give up trying to understand it.

sir no i will sure try to explain it in wide if u do not mind........ may u pl mention your email or whatsup no so that i ping u or email u the link ok

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3 minutes ago, mathspassion said:

sir no i will sure try to explain it in wide if u do not mind........ may u pl mention your email or whatsup no so that i ping u or email u the link ok

Of course not. Just post the steps here.

Just explain how you "count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square"

I assume this will just take 3 or 4 sentences.

Edited by Strange
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1 hour ago, Strange said:

Of course not. Just post the steps here.

Just explain how you "count the nos of dots(points where crossing with each others)start from origin to 3rd square  to 7th square"

I assume this will just take 3 or 4 sentences.

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1 hour ago, mathspassion said:

ok in few days I will quote a link(youtube) in which every thing will be clear ok

Don't bother. I don't watch videos. (They are the worst possible way of explaining things.)

If you can't explain it here in a few sentences, then this is obviously such an impossibly complicated process that it isn't worth spending any time on.

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If I'm not mistaken, this is a rather long assertion of the fact that for any N,

4 (N - 1) + (N - 2)^2 = N^2.

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17 minutes ago, uncool said:

If I'm not mistaken, this is a rather long assertion of the fact that for any N,

4 (N - 1) + (N - 2)^2 = N^2.

I assumed it was something along those lines, but couldn't work out what the OP was doing.

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2 hours ago, Strange said:

I assumed it was something along those lines, but couldn't work out what the OP was doing.

The idea was that the * the OP was using denoted the largest number gotten when labeling the vertices of a lattice inside a rectangle of the two given side lengths. So the 3rd square *3rd square denotes the largest number gotten when labelling the vertices of a lattice inside a 3 by 3 rectangle, as shown in the picture. But once we have that, it's not hard to see that that is equal to (A + 1)(B +1), where A and B are the "number of squares". So, translating the formula, we get N^2 = (N-1)(4) + (N-2)(N-2).

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20 hours ago, Strange said:

Don't bother. I don't watch videos. (They are the worst possible way of explaining things.)

If you can't explain it here in a few sentences, then this is obviously such an impossibly complicated process that it isn't worth spending any time on.

verrrry strange neither mention you mail nor want to see on youtube and saying it is very complicated process something wrong with you as you already mention in your first quote it is complicated process means you already have read example then you replied and in trolling quote you trying to say it is complicated complicated and complicated what is this ............dear Strang my suggestion is to you always appreciate others work .............do not try to down others. ok

19 hours ago, uncool said:

If I'm not mistaken, this is a rather long assertion of the fact that for any N,

4 (N - 1) + (N - 2)^2 = N^2.

first time i am seeing this formula i am not aware of it i was just working on X-Y AXIS  and got that results you know everything is related with each others if it is not we can not prove and proofs things ,ok     .

maths is the only subject where we use x y for example x+y=24 xy=12

ok ....take care

16 hours ago, uncool said:

The idea was that the * the OP was using denoted the largest number gotten when labeling the vertices of a lattice inside a rectangle of the two given side lengths. So the 3rd square *3rd square denotes the largest number gotten when labelling the vertices of a lattice inside a 3 by 3 rectangle, as shown in the picture. But once we have that, it's not hard to see that that is equal to (A + 1)(B +1), where A and B are the "number of squares". So, translating the formula, we get N^2 = (N-1)(4) + (N-2)(N-2).

can you prove this formula

N^2 = (N-1)(4) + (N-2)(N-2).

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14 hours ago, mathspassion said:

can you prove this formula

N^2 = (N-1)(4) + (N-2)(N-2).

Dude. Just multiply it out. This is like basic algebra. Post here if you need help, but if you just expand each term on the right hand side, you'll see the terms cancel and prove the identity.

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12 hours ago, Bignose said:

Dude. Just multiply it out. This is like basic algebra. Post here if you need help, but if you just expand each term on the right hand side, you'll see the terms cancel and prove the identity.

ok we know dear but there must a proof and prove ok.

Edited by mathspassion
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2 hours ago, mathspassion said:

ok we know dear but there must a proof and prove ok.

The proof is:

(N-1)(4) + (N-2)(N-2)

= (4N - 4) + (N-2)(N-2)

= (4N - 4) +  (N^2 - 4N + 4)

= 4N - 4N - 4 + 4 + N^2

= N^2

QED

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To see it more geometrically, you can see the (N - 2)^2 as a square in the center of the N^2 square. Then you can divide up the "border" into 4 parts. Each will have N - 1 points.

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17 hours ago, Strange said:

The proof is:

(N-1)(4) + (N-2)(N-2)

= (4N - 4) + (N-2)(N-2)

= (4N - 4) +  (N^2 - 4N + 4)

= 4N - 4N - 4 + 4 + N^2

= N^2

QED

dear it is very simple but do you know how this formula came into existence if not in the next quote i will define you i have already proved  through points .o.k wait for.......

16 hours ago, uncool said:

To see it more geometrically, you can see the (N - 2)^2 as a square in the center of the N^2 square. Then you can divide up the "border" into 4 parts. Each will have N - 1 points.

dear it is very simple but do you know how this formula came into existence if not in the next quote i will define you i have already proved  through points .o.k wait for.......i have done lot work on (more then 20 years).

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