# Having problems with math..here's why!

#### Matt C

88% of all math cannot be done.

HERE IS A OVERVIEW AS TO WHY:
Humans screwed up the entire system by making things up that the math system itself not only won't ever allow, but also makes it impossible for what people came up with, to actually be done. In other words, the math system itself prevents the way people are doing it, from being done. yet everyone thinks it's perfectly fine.

HERE IS AN IN DEPTH BREAKDOWN:

This is what people are taught:
1. This system is absolute
2. There are exceptions while saying it's still an absolute.

Ok that is the first major contradiction which in itself is a major major problem. Something that is an absolute, cannot have any exceptions to it at all for any reason, if there is an exception, then it's impossible to be an absolute. This cannot be both ways.

HERE IS AN EXAMPLE OF THOSE 2 ASPECTS PUT INTO PRACTICE:
The system (Addition) says that 2+3=5 and is absolute. one group of 2 units, and one group of 3 units.

MULTIPLICATION:
2x3=6 because there is either two groups of 3, or three groups of 2, and is an absolute. Increasing the amount of units from 5 to 6, makes this possible, otherwise there are only 5 units and the rules and system, of math outright prevent 2+3=6.

Multiplication outright requires there to be a total of 6 actual units. So it's either 2 groups of 3, or 3 groups of 2.

After teaching someone those basic concepts, they are ready to go but now start have a major difficult time because of BS like this.

EXAMPLE LAYOUT OF ONE OF THE MILLIONS OF FALLACIES IN MATH:
I will explain how attempting to obtain a 3rd variable in a problem that only contains 2 numbers is impossible because of the math system itself, and I will explain this Using height, width, and depth.

To begin let's say you want want to find the depth of a rectangular area that only has the numbers 3 and 2, and the 3 is on the top, and the 2 is on the side. So I have arranged this diagram showing the set up.
...
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The set up shows there are 3 across the top and 2 on the side, or 2 on the top and 3 on the side. But the main point that there are only 5 actual units, which will completely prevent this from being able to be done as a multiplication problem.

As explained above, there would have to be 6 actual units in the diagram (2 groups of 3 , or 3 groups of 2) which would make the diagram look like this:
...
.
.
.

The way the exception has this being done is impossible (Use one number twice) REALLY!!!. Doing that, now 2+3=6 because you would take the corner dot, and use it both across and down so you would have 3 on top, and 3 going down which makes now makes 6 with only 5 units... :shock:

[Moderator note: nobody had suggested the answer to your original question was 6, or that multiplication was an appropriate method for the question, so your objections are to things that didn't exist prior to your mention of them.]

So now a system that once stated 2+3=5 and is an absolute, is now saying Even though there are a total of 5 units, 2+3 is now going to be 6.
That's saying this circle is green, now the circle is blue in this case, without ever changing the circles color itself from green to blue.
I do not care who says what, or how many people say it, You cannot have both 3 on top and 3 on the bottom with a total of only 5 actual units.

So either
1. Basic math will completely prevent the problem above from being able to be done the way people are doing it.
OR
2. People doing the problem the way they are completely prevents basic math from being able to be done.

Either way you want to look at it, both sides would prevent the other from existing, which makes it impossible to do no matter what.

No rational human being will ever waste their time dealing with something that dumb and stupid. So teachers and tutor's and whomever else, really have a lot of nerve telling people they just don't want to learn of they're stubborn, after they got show a system that has 2+3=5 is absolute, then say 2+3=6 by manipulating the system totally contradicting themselves

I'll close with this:
For the people who are having problems with math, there is nothing wrong with you, and there isn't a problem with the system itself, the actual problem is people making things up in order to get the math system to fit into how they want, instead of actually applying the system itself to the problem.

There is no person on the face of this planet that is going to say what I have in this topic is a multiplication. The rules of addition says it's not, and so does multiplication.

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#### skipjack

Forum Staff
Your "88%" is an invented statistic. The so-called fallacy you give is based on a simple error. The top and side of your rectangle of dots share one dot, so the top and left side have 3 dots each, but can be shown, and would be counted, using just 3 + 2 = 5 dots instead of 2 Ã— 3 = 3 + 3 = 6.

The product of the width and length of a rectangle gives the area enclosed within the rectangle, not the number of "units" there are in just the top and left side of the rectangle, so what you objected to wouldn't be taught in school or suggested in any textbook and therefore shouldn't be regarded as a fallacy.

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#### Matt C

You just proved my point about people making up whatever they want instead of actually applying math rules and its system.

2 Ã— 3 = 3 + 3 = 6.
You cannot have that math equation without there there being 6 actual units present. you are making up numbers that do not exist, and somehow including them into the equation : :

Either Basic math prevents that from being done the way you're doing it, or the way your'e doing it says that basic math doesn't exist!
Either way you look at it, both ways cancel out the other.

There is only one group of 2, and one group of 3 which makes 5.

Multiplication also says it's not possible:
Only possible way for 6 to exist out of 2 and 3 is if there are either two groups of 3, or three groups of 2. which doesn't exist in this equation

There are only 5 units in this equation NOT 6 not 6, so it's either 3 on the top and 2 on the side, or it's 3 on the side and 2 on the top, you cannot use 3 on the top and 3 on the side without having 6 actual units.

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#### skipjack

Forum Staff
The multiplication by itself doesn't suffice because of the shared dot. Allowing for that, one has number of dots = 2 Ã— 3 - 1 = 3 + 3 - 1 = 5.

There is a corresponding error in your second diagram. Its left side has a total of 4 dots, one of which is shared with the top. Hence the total number of dots is 3 + 4 - 1 = 3 + 3 = 2 Ã— 3 = 6.

#### Micrm@ss

88% of math cannot be done? Well, aren't we lucky that 12% of math involves landing people on the Moon, sending a probe to Pluto, or designing computers and phone that you are using right now to post this message.

#### Matt C

your reply is exactly why people cannot do math, the basic math system and rules explain how the numbers work, and then here people come along completely contradicting the system and making things up instead of actually applying the math system to the problem.

The multiplication by itself doesn't suffice because of the shared dot. Allowing for that, one has number of dots = 2 Ã— 3 - 1 = 3 + 3 - 1 = 5.
under the Basic addition rules, 5 units are 5 and cannot ever be 6 without there actually being a 6th actual unit present. So the first diagram won't ever be 6 no matter what you attempt to do to it. Hence the reason for the second diagram.

There is a corresponding error in your second diagram. Its left side has a total of 4 dots, one of which is shared with the top. Hence the total number of dots is 3 + 4 - 1 = 3 + 3 = 2 Ã— 3 = 6.
There are no shared dots, not sure why and how you keep ignoring the actual amount of units present which is what actually matters, not sure what weird, irrational, and nonsense thing you're doing with it.
The second diagram is actually 6, because there are 3 units in the top row, and there are also 3 units in the side row because there are 6 actual units present. Neither side shares anything each side is individual and separate from each other.

You are mysteriously somehow getting 6 units out of the first diagram despite there only being no more than 5 actual units.
Multiplication says 6 cannot exist So How in the world, under the rules of basic math, are you mathematically getting 6 out of the first diagram, when there are only a total of 5 units? You cannot remove basic math from the equation without saying basic doesn't exist, and you cannot say basic math is being used but get any number in the first diagram other than 5. What you've done has both sides canceling out the other.

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#### skipjack

Forum Staff
One can get 6 out of 5 (as you put it) by counting one of the dots twice. This allows the count to reach six when the number of distinct dots is only 5.

In a rectangle with m dots down the left side and n dots across the top (with one dot shared), where m and n are unequal whole numbers, the total number of distinct dots in the top and left side together can be calculated by evaluating m + n - 1, which in general wouldn't be done using multiplication.

Confusion was caused by considering a square with m dots down the left side and m dots along the top (with one dot shared), where m = 3, but referring to it as a rectangle. For this situation, the total number of distinct dots in the top and side together can be calculated by evaluating m + m - 1, which can be written as 2m - 1, allowing its evaluation by using multiplication by two. The number 2 is used as a multiplier because 2 lines of dots are being considered, not because 2 happens to be one less than m.

#### Matt C

One can get 6 out of 5 (as you put it) by counting one of the dots twice.
The rules of both basic addition and multiplication, literally prevents that concept from even being possible. Basic math rules state that the only possible way to get 6, is to actually have 6.

I'd like to see you pay for something that costs one dollar and fifty cents, then hand the cashier 5 quarters telling them count one of the quarters 2 times, so there will be a total of one dollar and fifty cents. Let me know when you wind up with six quarters doing that if so, make a video of it and show it happening.

You still will only have 5 quarters no matter what you say, same applies to this problem, nothing you say matters, the basic rules of math tell you that what you're doing is not possible.

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#### Monox D. I-Fly

To begin let's say you want want to find the depth of a rectangular area that only has the numbers 3 and 2, and the 3 is on the top, and the 2 is on the side. So I have arranged this diagram showing the set up.
...
.
.

The set up shows there are 3 across the top and 2 on the side, or 2 on the top and 3 on the side. But the main point that there are only 5 actual units, which will completely prevent this from being able to be done as a multiplication problem.

As explained above, there would have to be 6 actual units in the diagram (2 groups of 3 , or 3 groups of 2) which would make the diagram look like this:
...
.
.
.
This is what happens when you are asked to find an area of a rectangle but you misinterpret area as half the perimeter instead.

#### studiot

Does anybody know a good bullshit meter repair facility, because my meter just shot off the scale and broke. • 2 people