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Declension
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02 Jan 2013, 11:53 pm

MCalavera wrote:
How does the number line show that the answer must be two? What does it mean to subtract 1 from 3 via a number line?


I think that your idea is that "subtraction of a positive number" on the mechanical side translates to "removing bars" on the visual side. I don't think this is right.

It would be better to say that "subtraction of a positive number" on the mechanical side translates to "moving left" on the visual side.

To see this, you should allow the left number to be negative. For example, what is -5 - 7? The answer is -12. It is easy to see this if you think of it as "moving left by 7". But if you think of it as "removing 7 bars", it doesn't make sense at all.



ruveyn
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03 Jan 2013, 12:20 am

Declension wrote:
MCalavera wrote:
How does the number line show that the answer must be two? What does it mean to subtract 1 from 3 via a number line?


.


Subtracting a number is an operation. Negative means go back. Adding a number is also an operation. Adding means go forward.

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Logicurse
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04 Jan 2013, 9:41 pm

Declension wrote:
MCalavera wrote:
How does the number line show that the answer must be two? What does it mean to subtract 1 from 3 via a number line?


I think that your idea is that "subtraction of a positive number" on the mechanical side translates to "removing bars" on the visual side. I don't think this is right.

It would be better to say that "subtraction of a positive number" on the mechanical side translates to "moving left" on the visual side.

To see this, you should allow the left number to be negative. For example, what is -5 - 7? The answer is -12. It is easy to see this if you think of it as "moving left by 7". But if you think of it as "removing 7 bars", it doesn't make sense at all.


In addition to that motion on the visual side will aid in understanding co-ordinate systems better.

I am now interested in what abstract concept corresponds to the extension of "removing bars" to 2 dimensional co-ordinate systems for a finite space.



JBlitzen
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05 Jan 2013, 3:29 pm

Why can't negatives be simply the number of more bars needed to fulfill subtraction after exhausting the actual supply? The deficit, call it.



ModusPonens
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05 Jan 2013, 7:14 pm

I see the origin of your confusion. The distance between two points, say 4 and 5, can be calculated as an integral on the interval [4,5]. It can be proved that if you take away a finite number of points of the interval where the integral is defined, it doesn't change its value. So the value is 1 either in (4,5) or [4,5] or (4,5] or [4,5).

This is closely related to the question "Is there a real number imediatly next to 1?" The answer is no. For example, 0.9999999... is equal to 1.



mereexistence
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05 Jan 2013, 10:42 pm

It sounds like we're talking about the "edge" of a defined region, and are having trouble visualizing whether that edge is lost in an act of subtraction. The answer is "no".

Zero is confusing to visualize because we think of it as a metaphysical starting point for anything tangible. Because of that, shift everything over by 1. So we have 1-2-3-4-5-6.

Now it seems to me that you're talking about the infinitely small edge between two numbers, so let's zoom in so we can see it. To the left of "5" is "4.9999999...". As soon as you get to "5.000000000", you're talking about "5". "4.999999...." is thought to belong to the "4" range.

So if you remove everything to the left of "5", it's nice and clean where the very edge of "5" is the beginning of your remainder.

The same can be said of zero on your number line... anything less than zero is before it, and it ranges up to "0.999999999...".



MCalavera
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06 Jan 2013, 11:01 am

ModusPonens wrote:
So the value is 1 either in (4,5) or [4,5] or (4,5] or [4,5).


This is new knowledge for me, so for that, I thank you. I still haven't figured it out perfectly, but this is definitely one step up the ladder.

I'll be studying some calculus online soon to see if this will eventually help me understand what's going on exactly.



Declension
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06 Jan 2013, 10:10 pm

JBlitzen wrote:
Why can't negatives be simply the number of more bars needed to fulfill subtraction after exhausting the actual supply? The deficit, call it.


Interesting! I never thought of it that way before. For me, it is always about "moving the cursor", not about using the number line as a resource. I guess this is why we need formal mathematics in order to communicate with each other - our mental pictures can sometimes be very different.



JBlitzen
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06 Jan 2013, 10:53 pm

Declension wrote:
JBlitzen wrote:
Why can't negatives be simply the number of more bars needed to fulfill subtraction after exhausting the actual supply? The deficit, call it.

Interesting! I never thought of it that way before. For me, it is always about "moving the cursor", not about using the number line as a resource. I guess this is why we need formal mathematics in order to communicate with each other - our mental pictures can sometimes be very different.

Well said. I think that's what I find really cool about this thread; seeing everyone's different mental pictures of the same basic thing.

Most message boards would be too busy sneering at the question to discover that.



Kenjuudo
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07 Jan 2013, 2:54 pm

Think of a pointer moving on the number line!

Code:
What is 5 - 4 ?

We start with 5:

                                  V
   ===|===|===|===|===|===|===|===|===|===|
     -2  -1   0   1   2   3   4   5   6   7

Then we move 4 left (because of subtraction):

                              V
1) ===|===|===|===|===|===|===|===|===|===|
     -2  -1   0   1   2   3   4   5   6   7

                          V
2) ===|===|===|===|===|===|===|===|===|===|
     -2  -1   0   1   2   3   4   5   6   7

                      V
3) ===|===|===|===|===|===|===|===|===|===|
     -2  -1   0   1   2   3   4   5   6   7

                  V
4) ===|===|===|===|===|===|===|===|===|===|
     -2  -1   0   1   2   3   4   5   6   7

We ended on 1. Therefore, 5 - 4 = 1.
Code:
What is -7 - 5 ?

We start with -7:

                                  V
   ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

Then we move 5 left (subtraction):

                              V
1) ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

                          V
2) ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

                      V
3) ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

                  V
4) ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

              V
5) ===|===|===|===|===|===|===|===|===|===|
    -14 -13 -12 -11 -10  -9  -8  -7  -6  -5

We ended on -12. Therefore, -7 - 5 = -12.


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