Negative square

In2infinity

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Hello there.

I was looking for an answer to a very simple question that is playing on my mind.

I can make a cross to form an x ,y axis, with zero at the center. I can label each half of the number line with positive or negative opposites. If I perform a calculation such as +2x * +3y I get a rectangle in the ++ space. And if I calculate -2x * -3y I get a rectangle in the - - space.

By when I try to explore this on my calculator it never shows me the right answer. As -x * -y seems to equal a positive value?

Why?
 
What would be your favorite way to assign signs to your rectangles? Why?
 
This way. As it allows me to assign the values in the correct numerical space. Zero sits at the center of the cross.

btw: The bottom left section is (a)
 

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Hello there.

I was looking for an answer to a very simple question that is playing on my mind.

I can make a cross to form an x ,y axis, with zero at the center. I can label each half of the number line with positive or negative opposites. If I perform a calculation such as +2x * +3y I get a rectangle in the ++ space. And if I calculate -2x * -3y I get a rectangle in the - - space.

By when I try to explore this on my calculator it never shows me the right answer. As -x * -y seems to equal a positive value?

Why?
Multiplying two negative numbers together does result in a positive answer; your calculator is giving you the "right answer".

Where did you get the picture you have posted?
(It looks like you may have misunderstood what was on that website.)
 
This way. As it allows me to assign the values in the correct numerical space. Zero sits at the center of the cross.

btw: The bottom left section is (a)
There are two signs in each square, but the multiplication results requires only one. which one's would you choose and why?
 
Not to beat on a dead horse, but: the fact that the product of two negatives is positive can be deduced from distributivity of the (ring of) real numbers.
 
Hi there. Thanks for your great feedback. But I still don't understand why two - - = +. In the image i shared, (sorry I forgot the website), But it was about mapping different types of square number spaces into four quadrants. For Example:

+x2 * +y2 = ++4 (top left)
-x2 * +y2 = - + 4 (top right)
-x2 * -y2 = - - 4 (bottom left)
+x2 * -y2 = + - 4 (bottom right)

This seems to make sense to me.

But what is the reason why on my calculator the operation is different. What is actually going on. I did not really understand the idea of "distributivity of the (ring of) real numbers". What does that mean??
 
But I still don't understand why two - - = +.
Are you asking why negative times negative is positive?
If so, this thread can be helpful:
 
Thank you for the link. I did spend some time going though all the points made, however, it still has not answered my question. In fact is raises more questions. Like this:


This is not something mathematicians "agreed on", and it can be proven.” -
If it can be proven then why don’t mathematicians agree?



And then this:

Start with this fact: -(-a)) = a. In other words, the opposite of the opposite of a number is the original number.

But this is not a FACT of the presentation of number I am considering. Negative numbers are at both 180 and 90 degree angles.

Furthermore there is a mathematical equation that is offered as proof:

x = ab + (-a)(b) + (-a)(-b).

On one hand

x = ab + (-a)[ (b) + (-b) ] (factor out -a from the last three terms)

= ab + (-a) * 0

= ab + 0

= ab.

But on the other hand,

x = [ a + (-a) ]b + (-a)(-b) (factor out b from the first two terms)

= 0 * b + (-a)(-b)

= 0 + (-a)(-b)

= (-a)(-b).

But if x = ab and x = (-a)(-b), then ab = (-a)(-b).


Yet when I look at this and compare it to the axis in the picture, a and b can be construed as the two different number lines. -a and -b found in the conclusion is actually the bottom left square???

The reason I am asking about this is because I am trying to build a calculator based on this number cross from the picture, but I keep getting the wrong answers. I am a fan of geometry, and it is easier to construct geometric ratio which can be compared in this way.

If anyone can actually point to a proof that is compatible with this ‘number cross’ model. After all it is only a simple x,y axis.

Thanks again for your valuable help.
 
I'd like to know what you mean by this?
Distributivity: [imath]a \times (b + c) = a\times b + a\times c[/imath]. Also, we define [imath]-a[/imath] as given by the equation [imath](-a) + a = 0[/imath]

[imath]a \times 0 = 0[/imath] for all [imath]a[/imath]. Indeed: [imath]a\times 0 = a \times (1-1) = a - a = 0[/imath]

[imath]a \times (-b) = - (a\times b)[/imath]. Indeed: [imath]a\times (-b) = a\times (0-b) = a\times 0 - a\times b = - (a\times b)[/imath]

[imath](-a)\times (-b) = a\times b[/imath]. Indeed: [imath]0 = (a + (-a))\times (-b) = a\times (-b) + (-a)\times (-b) = -(a\times b) + (-a)\times(-b)[/imath]. But since [imath]0 = -(a\times b) + (-a)\times(-b)[/imath] and [imath]0 = -(a\times b) + a\times b[/imath] we can deduce that [imath]a\times b = (-a)\times(-b)[/imath].
 
You are getting quite sophisticated answers to a simple question.

You are drawing rectangles in a coordinate plane and calculating their areas. Where you place the rectangles does not affect their area.

A different way to say this is to give a rule on how to measure length in a coordinate plane.

[math]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.[/math]
The square root function is defined as never negative. So all your examples involve multiplying positive numbers and getting a positive result. Nothing weird there. You are improperly assigning signs to length and width. What you are doing is computing lengths and widths differently in different quadrants. Your problem goes away if use a consistent method of measurement.
 
If it can be proven then why don’t mathematicians agree?
Which mathematician(s) disagree? I'm interested.

Yet when I look at this and compare it to the axis in the picture, a and b can be construed as the two different number lines. -a and -b found in the conclusion is actually the bottom left square???
Yes, but did you also consider the red arrow in your picture?
 
Area is always positive in the real world. Therefore response#12 might be the best way for you (OP) to think about this.

However, in calculus (more advanced maths) area can be measured negatively. And it kind-of fits in with the diagram that you posted. Therefore here's a very quick overview...

Think of an area as the sum of many vertical strips. Each strip is measured positively above the x-axis and negatively below. These strips are indicated by green arrows (and text) in the following diagram. If the strips are summed in a positive x direction (from left to right) then each strip is added to a running total of the area. If the strips are gathered right to left (in a negative x direction) then each strip is subtracted from a running total of area (x direction is shown with red arrows and text). This diagram shows how 4 areas can be calculated by adding/ subtracting vertical strips according to these sign conventions...

area.png

This is a very simplified, high level, view of how calculus finds an area.
 
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Distributivity: [imath]a \times (b + c) = a\times b + a\times c[/imath]. Also, we define [imath]-a[/imath] as given by the equation [imath](-a) + a = 0[/imath]

[imath]a \times 0 = 0[/imath] for all [imath]a[/imath]. Indeed: [imath]a\times 0 = a \times (1-1) = a - a = 0[/imath]

[imath]a \times (-b) = - (a\times b)[/imath]. Indeed: [imath]a\times (-b) = a\times (0-b) = a\times 0 - a\times b = - (a\times b)[/imath]

[imath](-a)\times (-b) = a\times b[/imath]. Indeed: [imath]0 = (a + (-a))\times (-b) = a\times (-b) + (-a)\times (-b) = -(a\times b) + (-a)\times(-b)[/imath]. But since [imath]0 = -(a\times b) + (-a)\times(-b)[/imath] and [imath]0 = -(a\times b) + a\times b[/imath] we can deduce that [imath]a\times b = (-a)\times(-b)[/imath].
Thanks for that. A great detail explanation. I will ponder this in more detail :)
 
You are getting quite sophisticated answers to a simple question.

You are drawing rectangles in a coordinate plane and calculating their areas. Where you place the rectangles does not affect their area.

A different way to say this is to give a rule on how to measure length in a coordinate plane.

[math]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.[/math]
The square root function is defined as never negative. So all your examples involve multiplying positive numbers and getting a positive result. Nothing weird there. You are improperly assigning signs to length and width. What you are doing is computing lengths and widths differently in different quadrants. Your problem goes away if use a consistent method of measurement.
Right so we root and square the function. Wow! I never thought of that!
 
Area is always positive in the real world. Therefore response#12 might be the best way for you (OP) to think about this.

However, in calculus (more advanced maths) area can be measured negatively. And it kind-of fits in with the diagram that you posted. Therefore here's a very quick overview...

Think of an area as the sum of many vertical strips. Each strip is measured positively above the x-axis and negatively below. These strips are indicated by green arrows (and text) in the following diagram. If the strips are summed in a positive x direction (from left to right) then each strip is added to a running total of the area. If the strips are gathered right to left (in a negative x direction) then each strip is subtracted from a running total of area (x direction is shown with red arrows and text). This diagram shows how 4 areas can be calculated by adding/ subtracting vertical strips according to these sign conventions...

View attachment 32295

This is a very simplified, high level, view of how calculus finds an area.
Thank you for your simplified explanation. I think this is the solution, explained in #12 in a simple to understand way for someone like me. I see we have a pair of red arrows the extends left and right from zero. The top are the positive set and the bottom are the negative set. Same for the Green Arrows. This resolves the problem of - numbers being opposite side. Now the 'space' above the x line are positive values and the 'space below the x line are negative. Ah! Its all makes sense. Except one point. See Below

Still I like this answer the most, as it does at least provide a geometric reason for the concept.
 
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