My Math Forum > Math Having problems with math..here's why!

 Math General Math Forum - For general math related discussion and news

 July 19th, 2018, 11:40 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 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 × 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. Last edited by skipjack; August 27th, 2018 at 10:20 AM.
July 19th, 2018, 11:42 PM   #3
Banned Camp

Joined: Jul 2018
From: beverly hills

Posts: 15
Thanks: 0

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

Quote:
 Originally Posted by skipjack 2 × 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 :O_o:

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.

Last edited by skipjack; July 30th, 2018 at 11:28 PM.

 July 19th, 2018, 11:51 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 The multiplication by itself doesn't suffice because of the shared dot. Allowing for that, one has number of dots = 2 × 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 × 3 = 6.
 July 20th, 2018, 12:17 AM #5 Senior Member   Joined: Oct 2009 Posts: 867 Thanks: 330 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.
July 20th, 2018, 12:27 AM   #6
Banned Camp

Joined: Jul 2018
From: beverly hills

Posts: 15
Thanks: 0

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.

Quote:
 Originally Posted by skipjack The multiplication by itself doesn't suffice because of the shared dot. Allowing for that, one has number of dots = 2 × 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.

Quote:
 Originally Posted by skipjack 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 × 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.

Last edited by skipjack; July 30th, 2018 at 08:45 PM.

 July 20th, 2018, 12:32 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 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.
July 20th, 2018, 12:49 AM   #8
Banned Camp

Joined: Jul 2018
From: beverly hills

Posts: 15
Thanks: 0

Quote:
 Originally Posted by skipjack 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.

Last edited by skipjack; January 16th, 2019 at 11:14 PM.

July 20th, 2018, 01:00 AM   #9
Senior Member

Joined: Nov 2010
From: Indonesia

Posts: 2,001
Thanks: 132

Math Focus: Trigonometry
Quote:
 Originally Posted by Matt C 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.

 July 20th, 2018, 01:16 AM #10 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 Does anybody know a good bullshit meter repair facility, because my meter just shot off the scale and broke. Thanks from jonah and Joppy

 Tags mathhere, problems

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post claudia_lovez_u Elementary Math 1 May 14th, 2012 07:24 PM jhunt47 Algebra 10 November 16th, 2009 11:22 PM Climaxx Computer Science 2 May 23rd, 2009 09:46 AM doomguardian Algebra 2 October 17th, 2007 01:56 PM mathhelp Algebra 12 May 24th, 2007 04:57 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top