## Math without even numbers?

For the discussion of math. Duh.

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liberonscien
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### Math without even numbers?

What would math be like if 1+1=3 and 3-1=1?
7x7=59
Basically, if two objects are introduced to each other, this causes a third object to be generated.

In this universe, even is only hypothetical and can't exist in real life.
Humans would have 3 legs, 3 eyes, or only one. So on and so forth.
Last edited by liberonscien on Fri Mar 25, 2016 1:11 am UTC, edited 2 times in total.

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### Re: Math without even numbers?

It's the Chocolate Chip thing, the mods are ruining the posts (hence why it now says fence post tree). It'll be gone by April 2nd.
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### Re: Math without even numbers?

liberonscien wrote:What would math be like if 1+1=3 and 3-1=1 and 3-2=0?
Basically, if two objects are introduced to each other, this caused a third object to be generated.
Edit:
What's up with my post?
I know I posted 2, the word, but the forum replaced it with 7, the word.

Huh, it is fixed.

Anyways, I figure that even numbers wouldn't really exist in this world.

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### Re: Math without even numbers?

If even numbers don't exist, what's the 2 doing in "3-2=0"?
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liberonscien
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### Re: Math without even numbers?

gmalivuk wrote:If even numbers don't exist, what's the 2 doing in "3-2=0"?

Touche.

Hmm.

I'll get rid of that.

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### Re: Math without even numbers?

0 is an even number as well. What is 1-1 supposed to be, or is subtraction even going to be defined at all? (Without zero, you don't have an additive identity, so it seems silly to speak of an additive inverse.)

chridd
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### Re: Math without even numbers?

Well, with 1+1=3, my first impression is that it's just relabelling the numbers... calling 1 "1", calling 2 "3", calling 3 "5", etc. Which doesn't change much; it's base 5 instead of base 10, but there are less confusing ways to write base 5. (Also how would you write 0? If it's written as -1, then - means something different...)

If you want instead to keep the results of operations the same, then you can no longer add anything, because odd + odd = even. You could still multiply just as you could before, because odd × odd = odd; however, I believe that's isomorphic to ordinary integer multiplication, so it would still be relabelling numbers, and also removing a useful operation. (Although I guess you could make addition a ternary operation, so 1+1 is not a well-formed expression but 1+1+1 is... or an any-odd-number–ary expression) On the other hand, if you instead removed the odd numbers, the even numbers are closed under both addition and multiplication, and multiplication would be different since there's no 1.

If you always round either up or down to an odd number, then addition can work (assuming you had a consistent rule for which direction to round), but if you always rounded in the same direction, I believe it's still equivalent to addition, but with relabelled numbers again. Addition's relationship to multiplication would be different, though; e.g., if you rounded up, 3+3=7, but there isn't anything you can multiply by 3 to get 7.

curiosityspoon wrote:0 is an even number as well. What is 1-1 supposed to be, or is subtraction even going to be defined at all? (Without zero, you don't have an additive identity, so it seems silly to speak of an additive inverse.)
You don't need to have an additive inverse to do subtraction, though; subtraction can still work e.g. in the natural numbers (though it's only defined for some combinations of numbers, then).

liberonscien wrote:Basically, if two objects are introduced to each other, this caused a third object to be generated.
What exactly do you mean by "introduced"? I mean, we can talk about two things, even if those two things have absolutely no connection with each other and aren't near each other.
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### Re: Math without even numbers?

Well, my first impression (based on the few examples in the OP) that we were defining a new "addition" over the odd numbers, which is their normal sum plus one. "Subtraction" would then be their normal difference minus one.

The resulting group would be isomorphic to normal addition over the integers, with -1 being the additive identity. Basically nothing interesting happens.

It's very similar to just doing math without the odd numbers, only using even numbers, and keeping addition/subtraction defined the same - everything behaves the same, the numbers are just twice as big. Then you just subtract 1 from every number, and redefine addition/subtraction so the mappings don't change.

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### Re: Math without even numbers?

Would 1-1=0 or -1? Is 0 an even number or an odd number?

Would you have to always deal with an odd number of figures? So you couldn't say 1+1, you can only add them together an odd number of times - 1+1+1? Are fractions and ratios also banned? Also you couldn't square numbers, you could only cube them or use the 5th power or so, so numbers would get very huge quite quickly. Nor could you use square roots. Pi might or not be banned, but no one knows yet.
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### Re: Math without even numbers?

0 is odd. 1+1=3, 3+3=7, 5+5=11, 7+7=15, 9+9=19, 11+11=33.
Addition always rounds up to the next odd number, even when the odd number is in the tens place as evidenced by 11+11=33.
Subtraction does the opposite. 9-3=5.

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### Re: Math without even numbers?

liberonscien wrote:0 is odd.
It definitely isn't.
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liberonscien
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### Re: Math without even numbers?

gmalivuk wrote:
liberonscien wrote:0 is odd.
It definitely isn't.

Then what is it?

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### Re: Math without even numbers?

It's even. That's the other possibility, after all.
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### Re: Math without even numbers?

Xanthir wrote:It's even. That's the other possibility, after all.

I meant, - Never mind.

I figure math is rather exponential this way.

chridd
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### Re: Math without even numbers?

So if you're going by digits (since you're disallowing e.g. 23), and treating 0 as odd, then does that mean that you're treating 10 (and 30, 50, 100, 37050, etc.) as odd?
(In which case, it's actually, math without numbers containing the digits 2, 4, 6, 8?)
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### Re: Math without even numbers?

chridd wrote:So if you're going by digits (since you're disallowing e.g. 23), and treating 0 as odd, then does that mean that you're treating 10 (and 30, 50, 100, 37050, etc.) as odd?
(In which case, it's actually, math without numbers containing the digits 2, 4, 6, 8?)

0 is even.

Basically, in this kind of math, the universe hates even numbers.

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### Re: Math without even numbers?

liberonscien wrote:Basically, in this kind of math, the universe hates even numbers.
Yet it still uses base 10 (an even base)?

In odd bases, though, there are numbers that odd but end in an even digit... on the other hand, in odd bases, if you have an odd number where all the digits are odd, then there's an odd number of digits in the number. (Although I'm not sure how the round-to-nearest-odd thing would work there.)
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### Re: Math without even numbers?

chridd wrote:
liberonscien wrote:Basically, in this kind of math, the universe hates even numbers.
Yet it still uses base 10 (an even base)?

In odd bases, though, there are numbers that odd but end in an even digit... on the other hand, in odd bases, if you have an odd number where all the digits are odd, then there's an odd number of digits in the number. (Although I'm not sure how the round-to-nearest-odd thing would work there.)

Yes, I think this universe also allows perpetual motion devices. I could be wrong though.

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### Re: Math without even numbers?

liberonscien wrote:0 is odd. 1+1=3, 3+3=7, 5+5=11, 7+7=15, 9+9=19, 11+11=33.
Addition always rounds up to the next odd number, even when the odd number is in the tens place as evidenced by 11+11=33.
Subtraction does the opposite. 9-3=5.

Still two operands though?
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### Re: Math without even numbers?

liberonscien wrote:0 is odd. 1+1=3, 3+3=7, 5+5=11, 7+7=15, 9+9=19, 11+11=33.
Addition always rounds up to the next odd number, even when the odd number is in the tens place as evidenced by 11+11=33.
Subtraction does the opposite. 9-3=5.

Still two operands though?

?

There is still multiplication and division.

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### Re: Math without even numbers?

So, we replace even numbers with the next odd number? So "11" is a three-digit number. We have five operations: Addition, subtraction, multiplication, and division. And when describing my dinner last night, I'd say "I nine an elevenderloin with a fivek"?
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### Re: Math without even numbers?

liberonscien wrote:
liberonscien wrote:0 is odd. 1+1=3, 3+3=7, 5+5=11, 7+7=15, 9+9=19, 11+11=33.
Addition always rounds up to the next odd number, even when the odd number is in the tens place as evidenced by 11+11=33.
Subtraction does the opposite. 9-3=5.

Still two operands though?

?

There is still multiplication and division.

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### Re: Math without even numbers?

gmalivuk wrote:
liberonscien wrote:0 is odd.
It definitely isn't.

If one can define math to only have odd numbers, 0 surely can be defined to be odd.
Or going the other way round: the symbol "0" represents the neutral element in respect to addition, so while that may be -1 in the notation of regular math, in the proposed system it could be 0. That's also very close to what most people consider a "real world" definition of 0.
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### Re: Math without even numbers?

So are you saying that 1+(-1)=1, while 1-1=-1?
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### Re: Math without even numbers?

How many are these? * * *
How many are these? * * * *
How many rows of asterisks are there?

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### Re: Math without even numbers?

ucim wrote:How many are these? * * *
How many are these? * * * *
How many rows of asterisks are there?

Jose

In our universe, when I see this:
* * *
* * * *

It stays this:
* * *
* * * *

In this hypothetical universe, when I see this:
* * *
* * * *

This is generated:
* * *
* * * * * * *
* * * * *

In this universe, perception matters. In this universe, if, by natural forces, two rocks are brought together and no one sees this happen, then nothing happens.
However, if two rocks are brought together, and someone sees this, then a third rock spontaneously comes into existence.

In our universe, if one sees * *, it stays * *.
In this hypothetical universe, if one sees * *, it generates another *, turning it into this * * *.

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### Re: Math without even numbers?

But 2 is really 1.9999 recurring anyway. Apart from pi and decimals divided by 7, 13 and so on (and possibly zero) aren't all numbers odd?
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### Re: Math without even numbers?

From the perspective of addition, this just looks like a straightforward base 5 system. However, multiplication no longer follows the pattern of "repeated addition" that we're used to. To use the initial example, 7x7 is 59, but 7+7+7+7+7+7+7 is only 55.

Also, how does subtraction work? If we take the assumption that all evens are "rounded up", then 5-3 is 3, but this violates the idea that subtraction is the inverse of addition since 3+3 is 7. We can fix this by letting -1 be the additive identity, and -3 be the additive inverse of 1, and -5 be the additive inverse of 3, etc.

Once you do that, though, why not just "add one" to all your symbols and work in a universe of only evens? Zero becomes zero again, and the negatives and positives work out as they ought. you still have a base 5 system that does the same job, but without weird consequences.

Edit: ok, there are still very weird consequences to multiplication and exponentiation if you assume 2x2 is still 4 - you no longer have a multiplicative identity.

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### Re: Math without even numbers?

Gwydion wrote:From the perspective of addition, this just looks like a straightforward base 5 system. However, multiplication no longer follows the pattern of "repeated addition" that we're used to. To use the initial example, 7x7 is 59, but 7+7+7+7+7+7+7 is only 55.

Also, how does subtraction work? If we take the assumption that all evens are "rounded up", then 5-3 is 3, but this violates the idea that subtraction is the inverse of addition since 3+3 is 7. We can fix this by letting -1 be the additive identity, and -3 be the additive inverse of 1, and -5 be the additive inverse of 3, etc.

Once you do that, though, why not just "add one" to all your symbols and work in a universe of only evens? Zero becomes zero again, and the negatives and positives work out as they ought. you still have a base 5 system that does the same job, but without weird consequences.

Edit: ok, there are still very weird consequences to multiplication and exponentiation if you assume 2x2 is still 4 - you no longer have a multiplicative identity.

5-3=1

Also, in additive things like addition and multiplication, all even numbers in the answer are bumped up by one.
In subtracting things like subtraction and division all even numbers are reduced by one.

5+5=10. 10 is bumped up to 11. So 5+5=11
11-5=6. 6 is bumped down to 5. 11-5=5
5x5 is 25. 2 gets bumped up to 3. So 5x5=35

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### Re: Math without even numbers?

liberonscien wrote:5x5 is 25. 2 gets bumped up to 3. So 5x5=35

Which means that multiplication doesn't satisfy our normal expectation, that multiplication is like repeated addition, and that it should distribute over addition.

5 = 1+1+1, so by distribution we'd expect 5x to be x+x+x, if it were to satisfy distribution... and 5+5+5 = 11+5 = 17, under your definition of addition. This would then match up with the base-5 mapping (5odds => 310; 3*3 = 910 = 145 => 17odds)

If we go with your definition, where 5*5 = 35, then we also can't have true division - 5*7 is also 35, which means that 35/5 is undefined... same way you can't divide by zero, with normal numbers.

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### Re: Math without even numbers?

chridd wrote:(Also how would you write 0? If it's written as -1, then - means something different...)

c.f. Two's Complement and One's Complement negatives in binary, but now rather than there being "two zeros" (in the latter, and less useful for using binary addition of a negative number to perform a binary subtraction), there could be none at all. Theoretically like just as there is no year 'zero' in 'proper' AD/CE or BC/BCE numbering systems.

I'm not sure if it'd be a useful mathematical system, but consider the following alternate 'integer-and-a-half' numbers on a number-line:

...-5.5, -4.5, -3.5, -2.5, -1.5, -0.5, +0.5, +1.5, +2.5, +3.5, +4.5, +5.5...

That's a regular list, moving up and down the line by one spot always adds or removes the same value (1) at every step, and yet there's no 'zero', it is skipped over. Imagine, instead, that each number is represented by its double:

...-11, -9, -7, -5, -3, -1, +1, +3, +5, +7, +9, +11

Maybe that's mathematically 'useful', in some way (even in ways that the "only on the halves" version is not). It doesn't have a 'zero' or any other even number, but for simple accounting purposes you might be able to restrict yourself to odd numbers only.

But if you did have two three-legged men and counted the legs... They'd have <the number between 5 and 7> legs in total, for which you'd be forced to have something (5.5, being the number half way between adjacent numbers 5 and 7?) to let you deal with that. And you'd surely get that enough times that it'd ruin your attempt to 'forget' that even numbers are a thing.

You'd have to quite explicitly contrive to never have such an 'imaginary' even number of things that all had 'real' odd numbers, or else you'd end up with an 'imaginary' total. (c.f. trying to deal with products of actual Imaginary numbers. An even number of Imaginary numbers, multiplied together, gives you a number within the Real set. It's easy to avoid encountering Imaginary numbers if you're using just Real numbers in everything that you do (save for actually forcing an Real-and-Even root of a Real-and-Negative number) , but under what system do you stop an Imaginary system from often manifesting back into the Real dimension..?

Really, if your society had an odd number of limbs with an odd number of digits upon them, you could quite easily have an odd-based number-system, but at any level of mathematical enquiry, there'd likely be need to know that a number is even. In an odd base using digits true to our own (albeit extended or truncated from our 0..9), you'd do something similar to1 the trick in working out whether a number is divisible by three. Add the parities of the digits of the numbers together2, and see if it's a 'special' number, however much you try to ignore the possibility of even numbers...

.....

1 But not identical. An explanation for why, with decimal numbers, you find that Sum(digits(number_divisible_by_3)) recurses down to 3, 6 or 9 and Sum(digits(number_divisible_by_9)) recurses down to only 9 is that 9 comes down to the base-number minus 1, and 3 is a (the only!) divisor of that number.

In Base 17dec (as a contrived example), numbers that are multiples of 16dec include 1F17 which deconstructed into decimal is 1x171 + 15x170, whose digits which sums to 16dec, with 2E17 (i.e. 2+14), 3D17 (i.e. 3+13), etc to follow. And numbers divisible by 4 in Base17 are G (16dec), 13, 17, 1B, 1F (as above), 22, 26, 2A, 2E (also as above), 31, etc, and digits sum to numbers that reduce to '4' numbers, and even numbers (those already listed, plus the like of 15) are also roots from Base-minus-one. (1517 digit-sums to 6, which is not 16, nor 4-divisible, but is 2-divisible. The next footnote also applies to this.)

I'm sure someone (if not in antiquity prior even to Pythagoras or Plato, definitely someone more modern) has already fully explained it all, because it's simple enough that I worked this all out myself almost as soon as I learnt about Hexadecimal. 3216 -> 3+2=5, is thus dividable by 5 (=50dec) and 11116 -> 1+1+1=3, whilst also being 273dec -> 2+7+3=12 -> 1+2=3. Also 15 (F16) works exactly as per 9 in decimal, but 9 in hexadecimal doesn't work so well (9, 18dec=1216 -> 1+2->3, 27dec=2A16 -> 2+(A16==10dec)=(C16=12dec) which is a multiple of 3 ... all obviously confirmable as 'divisible by three' with the hexadecimal digit-sums, but the 'divisible by nine' clue just isn't there.

And, with binary, you can only look for "multiples of 1". Which, amazingly, if you add up the digits of any binary number and then recurse the process (still in binary) until you get down to a single digit will give you 1 for every number... because every number (technically) has '1' as a factor. And every other base has '1' as a factor of the number that is 'Base-minus-one', as well, for completeness.
/1

2 Even+even=even ('0'2=even, 22=even); even+odd=odd ('0'3=odd, 23=odd); odd+even=odd (12=odd, 32=odd), odd+odd=even (11=even, 33=even); X+|Y+Z|=f{X,f{Y,Z}} (123=1+(even+odd)=1+odd=odd+odd=even)

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### Re: Math without even numbers?

Just like introducing two rocks to each other would cause the universe to generate a third rock, the same would occur for living things.

Also, in this system, there wouldn't be a sex binary, but a sex trinary.
Just throwing that out there.
Also, trinary is a word, right?

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### Re: Math without even numbers?

liberonscien wrote:Also, in this system, there wouldn't be a sex binary, but a sex trinary.
Just throwing that out there.
Also, trinary is a word, right?

Not only is trinary a word, you should look up plant sexual reproduction. It's... well, not even.

stopmadnessnow wrote:Apart from pi and decimals divided by 7, 13 and so on (and possibly zero) aren't all numbers odd?
How about normal vectors? Or even Normal numbers? Rumor has it that abnormal numbers are odd ducks indeed.

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### Re: Math without even numbers?

ucim wrote:
liberonscien wrote:Also, in this system, there wouldn't be a sex binary, but a sex trinary.
Just throwing that out there.
Also, trinary is a word, right?

Not only is trinary a word, you should look up plant sexual reproduction. It's... well, not even.

stopmadnessnow wrote:Apart from pi and decimals divided by 7, 13 and so on (and possibly zero) aren't all numbers odd?
How about normal vectors? Or even Normal numbers? Rumor has it that abnormal numbers are odd ducks indeed.

Jose

It is almost even and odd.

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### Re: Math without even numbers?

Even with nearly all numbers being 1.999999... and so on? And Jose refers to numbers (like e) where the last digit is not known.
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### Re: Math without even numbers?

ucim wrote:Not only is trinary a word,

The standard word is "ternary" actually (but Wikipedia acknowledges "trinary" enough to redirect to ternary).

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### Re: Math without even numbers?

Derek wrote:
ucim wrote:Not only is trinary a word,

The standard word is "ternary" actually (but Wikipedia acknowledges "trinary" enough to redirect to ternary).

So, you're telling me that even trinary is an odd word?

Jose
Order of the Sillies, Honoris Causam - bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith - bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me - you really made a difference.

Mike Rosoft
Posts: 63
Joined: Mon Jun 15, 2009 9:56 pm UTC
Location: Prague, Czech Republic

### Re: Math without even numbers?

Careful. If you can't have even numbers, what's 1+1? It's a term with *two* numbers, but even numbers don't exist! So you can't have 1+1, you need to use 1+1+1. Now you might want to say 1+1+1=3 ... except that you can't, because you'd get a formula with *four* numbers!

I don't think that you can get this to work without running into a paradox.

cyanyoshi
Posts: 407
Joined: Thu Sep 23, 2010 3:30 am UTC

### Re: Math without even numbers?

Mike Rosoft wrote:Careful. If you can't have even numbers, what's 1+1? It's a term with *two* numbers, but even numbers don't exist! So you can't have 1+1, you need to use 1+1+1. Now you might want to say 1+1+1=3 ... except that you can't, because you'd get a formula with *four* numbers!

I don't think that you can get this to work without running into a paradox.

We could get around the problem of sums by leaving any sums with a not-odd number of terms undefined. So sum(1,1,1)=5, but sum(1,1) wouldn't be allowed. It's like how 1/0 and (-2)^(1/2) can't be evaluated if you restrict yourself to just the real numbers. What's the problem here?

elasto
Posts: 3751
Joined: Mon May 10, 2010 1:53 am UTC

### Re: Math without even numbers?

cyanyoshi wrote:We could get around the problem of sums by leaving any sums with a not-odd number of terms undefined. So sum(1,1,1)=5, but sum(1,1) wouldn't be allowed. It's like how 1/0 and (-2)^(1/2) can't be evaluated if you restrict yourself to just the real numbers. What's the problem here?

They are thinking the problem is that you can't do 1+1+1=3 because it has a total of four numbers; But it's ok - it just contains (3+1) numbers - whatever that is