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# The deconstruction and reconstruction of numbers pt. 2

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Posted 25 May 2013 - 11:13 PM

From now on, the "+" symbol no longer means to combine the count of objects (e.g. 3 things plus 4 things make 7 total things would not be modeled as 3 + 4 = 7).

Instead, the use of the + symbol is to show  5 + 3 = 7 to resolve the question of "how many spaces between objects are created when you line up five things and three things"

(ex: t _ t _ t _ t _ t _ t _ t _ t ) and this is now to be considered addition

If the other basic computational symbols maintained the same relationship to addition as they had before this new convention what would be the answers to the following problems?

4 - 3 = ?

3 x 3 = ?

9 / 3 = ?

sqrt (36) = ?

Edited by BMAD, 25 May 2013 - 11:15 PM.

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### #2 austinm

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Posted 26 May 2013 - 02:14 AM

Notation: I use " +' ", " -' ", " x' ", and " /' " to represent the four arithmetic operations.

I interpret your definition of +' as: a +' b = append b items to an a-list and count interstices. (N.B. it's an operator definition, and that the identity element for +' is 1.)

1.

If -' is defined as a -' b = remove b items from an a-list and count interstices (i.e. the logical complement of +'), then

Spoiler for 4 -' 3 =

If a -' b is defined as the inverse of the +' operation, such that a +' b -' b = a = a -' b +' b, then

Spoiler for 4 -' 3 =

(Using Common Notion 2: "If equals be added to equals, the wholes are equal.")

Note that the second comports with the notion that a -' a should equal the identity element: 1. (This is why I favor the second.)

2.

If x' is defined as repeated addition (via +'), such that 2 x' 4 = 4 +'4, then

Spoiler for 3 x' 3 =

Note that the ordering is that implied by the old-fashioned reading of 2 x 4 as "twice four," or "two fours." This definition has the interesting, and possible undesirable consequence that 1 is a left-identity but not a right-identity: 1 x' 4 = 4, but 4 x' 1 = 3. (All multiplications not containing a one commute.)

Lack of universal commutivity can be remedied by defining 1 x' 4 as 4 +' 0; i.e. multiplying by one is adding nothing to 4, which then requires the final counting of interstices. In this case 1 x' 4 = 3 = 4 x' 1, all multiplication commutes, but there is no multiplicative identity.

I'd be interested to hear opinions on which definition of 1 x' a is preferable. The second, containing some appeal, leads to a breakdown in distributivity when 1's are involved. The first, I believe, allows for distributivity in all cases. (I've not rigorously proven it, though.)

3.

Properly speaking, I'd say

Spoiler for the answer has to be...

Spoiler for That's because

However, we don't really want that result, do we? So, let's clean up our definitions by dropping the notion of items-and-spaces, and just go with

a +' b = a + b -1 (where " + " is the usual addition operator),

a x' b = a x b -1 (ditto).

Note that a +' 1 = a, and a +' b = b +' a. Good so far.

a x' b = b x' a, but we still have no (and have no hope of a) multiplicitave identity element. This has grave consequences for the idea of division, but we forge ahead blindly...

Defining division as the inverse of the multiplication operation, such that a x' b /' b = a = a /' b x' b, we can use the above to determine that

Spoiler for 9 /' 3 =

4.

Using the most-recent definition of " x' ", and assuming that sqrt'(a) is the inverse of a x' a, we see that

Spoiler for sqrt'(36) =

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### #3 phil1882

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Posted 26 May 2013 - 02:14 AM

Spoiler for

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Posted 26 May 2013 - 02:59 AM

Notation: I use " +' ", " -' ", " x' ", and " /' " to represent the four arithmetic operations.

I interpret your definition of +' as: a +' b = append b items to an a-list and count interstices. (N.B. it's an operator definition, and that the identity element for +' is 1.)

1.

If -' is defined as a -' b = remove b items from an a-list and count interstices (i.e. the logical complement of +'), then

Spoiler for 4 -' 3 =

If a -' b is defined as the inverse of the +' operation, such that a +' b -' b = a = a -' b +' b, then

Spoiler for 4 -' 3 =

(Using Common Notion 2: "If equals be added to equals, the wholes are equal.")

Note that the second comports with the notion that a -' a should equal the identity element: 1. (This is why I favor the second.)

2.

If x' is defined as repeated addition (via +'), such that 2 x' 4 = 4 +'4, then

Spoiler for 3 x' 3 =

Note that the ordering is that implied by the old-fashioned reading of 2 x 4 as "twice four," or "two fours." This definition has the interesting, and possible undesirable consequence that 1 is a left-identity but not a right-identity: 1 x' 4 = 4, but 4 x' 1 = 3. (All multiplications not containing a one commute.)

Lack of universal commutivity can be remedied by defining 1 x' 4 as 4 +' 0; i.e. multiplying by one is adding nothing to 4, which then requires the final counting of interstices. In this case 1 x' 4 = 3 = 4 x' 1, all multiplication commutes, but there is no multiplicative identity.

I'd be interested to hear opinions on which definition of 1 x' a is preferable. The second, containing some appeal, leads to a breakdown in distributivity when 1's are involved. The first, I believe, allows for distributivity in all cases. (I've not rigorously proven it, though.)

3.

Properly speaking, I'd say

Spoiler for the answer has to be...

Spoiler for That's because

However, we don't really want that result, do we? So, let's clean up our definitions by dropping the notion of items-and-spaces, and just go with

a +' b = a + b -1 (where " + " is the usual addition operator),

a x' b = a x b -1 (ditto).

Note that a +' 1 = a, and a +' b = b +' a. Good so far.

a x' b = b x' a, but we still have no (and have no hope of a) multiplicitave identity element. This has grave consequences for the idea of division, but we forge ahead blindly...

Defining division as the inverse of the multiplication operation, such that a x' b /' b = a = a /' b x' b, we can use the above to determine that

Spoiler for 9 /' 3 =

4.

Using the most-recent definition of " x' ", and assuming that sqrt'(a) is the inverse of a x' a, we see that

Spoiler for sqrt'(36) =

Very well thought out answer.  I am curious as to how you resolve the following identity for square roots

This is using the traditional sense of computation symbols

b=  a + 1

b^2 = a^2 +2a+1

sqRoot(b^2)=sqRoot(a^2+2a+1)

sqRoot(b^2)=sqRoot((a+1)^2)

Using the new definitions of the number system with the above identity

36 = (a+1)^2  how would the solution proceed?

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### #5 austinm

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Posted 26 May 2013 - 04:18 AM

Very well thought out answer.  I am curious as to how you resolve the following identity for square roots

This is using the traditional sense of computation symbols

b=  a + 1

b^2 = a^2 +2a+1

sqRoot(b^2)=sqRoot(a^2+2a+1)

sqRoot(b^2)=sqRoot((a+1)^2)

Using the new definitions of the number system with the above identity

36 = (a+1)^2  how would the solution proceed?

(On a computer impeded by broken keyboard now--please excuse necessarily-brief response.)

(Letters seven, seven-plus-one, and (crucially!) lowercase-quote do not work!)

I.e. 36 = (a plus-prime 1) times-prime (a plus-prime 1) = a times-prime a.

Suppose one wants to multiply out (a plus-prime 1) times-prime (a plus-prime 1) (and assume my instinct is correct about ability to distribute) we immediately run into still-open question about definition of multiplication by and of one. Will respond later at time possessed of full I/O.

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