Division is not absolute?

General Questions
#1

A guy on another message board I frequent thinks he can prove that division isn’t absolute. Now, my math skills literally don’t extend past what I had in high school, and I barely understand what he’s driving at, but I told him I’d post a link to his ‘proof’ here, where I know there are many, many math-inclined people.

Here’s what he wrote. I’d ask that, for his sake, you don’t pay attention to the pretty terrible grammar. Just tear apart (or support) his idea as necessary.

I’ll direct him to this thread, or you can email the dude at bioteq@gmail.com.

#2

2[:rolleyes:[sup]10[/sup]]

You lost me at 25/5 = 32
At first it looked like a severe place-value mistake. He has 35 = 15 and 25 = 10, both correct. But they should be in the same place. In other words, after using 3 and coming out with 10 left, he should have added 2 to that 3, not put it after the 3, which gives you the answer you know is right: 5.

So it’s some kind of mistake, apparently. But then he starts mumbling about pi, and using a whole bunch of terms without defining them, or indeed without making any sense at all. Sounds like a huge heaping load of BS to me. Can anyone just translate this into English perhaps?

#3

It looks like nonsense to me.

What he seems to be saying is this.

Conventionally, we solve 25/5 like this. We multiply 5 five times (and write down “5” to remind ourselves that this is what we have done. We then take that away from 25 and find – presto! – that the remainder is zero. We conclude that 5 goes into 25 five times.

Alternatively, we can solve 25/5 like this. We multiply 5 three times, and write down “3” to remind ourselves that that is what we have done. We then take that away from 25 and find that the remainder is 10. We then multiply 5 two times and write down “2”. We take that away from 10, and the remainder is zero. We have finished our division. We find that we have written down “3” and “2” – 32!

But of course 3 and 2 does not equal 32. 3 and 2 equals five. We have taken 5 away from 25 five times, not 32 times.

Either your friend is an idiot, or this is a laboured and badly-written joke.

#4

He is saying that 5 goes into 25 five times with a remainder of zero. Fine.

His other division is crackpot. He says 5 goes into 25 thirty times with a remainder of 10. Then 5 goes into 10 twice with a remainder of zero. Ergo 32. The flaw is 5 goes into 25 thirty times with a remainder of -125, not 10.

#5

I pointed it to this thread and here’s what he sent me in an email.

Thanks, guys.

#6

My favorite along those lines is 64/16.

Just cross out those pesky sixes and the answer is right before you. Sure we can extend this to apply to all numbers :confused:

#7

What this guy has done is to misalign the numbers while performing long division. He wrote:


   32
   __
5 )25
  -15
   --
   10
  -10
   --
    0

The 3 is obviously in the tens place, in which case the 15 should actually be 150. Since you cannot subtract 150 from 25 without going negative, the placement of the 3 is invalid, and the example falls apart. Of course, you could try to put the 3 in the ones place and see what happens:


    3.(20)  <--- [3] Note that this represents 20 tenths.
   ____
5 )25.0
  -15 |  <--- [1] Don't forget to drop down the zero!
   -- v
   100
  -100  <--- [2] You need a 20 for this step, not a 2!
   --
    0

So now you have 3 and 20/10, more commonly known as 5! Looks like division is absolute after all.

#8

What he’s doing in the 25/5=32 thing isn’t division. It sort of looks like division, but it’s some other algorithm that doesn’t give you the right answer to 25/5. Applying his algorithm to 25/32 will not give you 5, I guarantee. He sounds like a very interesting individual.

#9

I assume we’re just talking about positive integer division here. If so, let’s take a minute to prove that it works.

Take any two integers a and b with a > b. We want to prove that there are integers q and r, 0 < r < b, such a = qb + r. Obviously, a = nb + (a - nb) for any integer n. Now, the integers are Archimedean, which means that for any integers a and b, there is an integer n such that a < nb. This implies that there is a least such n, which we’ll denote [symbol]n[/symbol]. In that case, a = [symbol]n[/symbol]b + (a - [symbol]n[/symbol]b), and a - [symbol]n[/symbol]b < b. If not, a > ([symbol]n[/symbol] + 1)b, which contradicts our assumption that a < [symbol]n[/symbol]b.

So that’s the existence proof. Now, let’s assume that for some pair of integers (a, b) (a > b) that there are two sets of integers, (q[sub]1[/sub], r[sub]1[/sub]) and (q[sub]2[/sub], r[sub]2[/sub]), such that a = q[sub]1[/sub]b + r[sub]1[/sub], a = q[sub]2[/sub]b + r[sub]2[/sub], 0 < r[sub]1[/sub] < b, and 0 < r[sub]2[/sub] < b. Can this be?

I think it’s obvious that q[sub]1[/sub]b + r[sub]1[/sub] - (q[sub]2[/sub]b + r[sub]2[/sub]) = 0, or (q[sub]1[/sub] - q[sub]2[/sub])b + (r[sub]1[/sub] - r[sub]2[/sub]) = 0. This implies that (q[sub]1[/sub] - q[sub]2[/sub])b = r[sub]2[/sub] - r[sub]1[/sub].

Without losing generality, we can assume that r[sub]1[/sub] < r[sub]2[/sub], which implies that 0 < r[sub]1[/sub] < r[sub]2[/sub] < b, which implies that 0 < r[sub]2[/sub] - r[sub]1[/sub] < b - r[sub]1[/sub] < b.

With r[sub]2[/sub] - r[sub]1[/sub] > 0, it follows that q[sub]1[/sub] - q[sub]2[/sub] > 0. We can assume that q[sub]1[/sub] != q[sub]2[/sub], which implies that q[sub]1[/sub] - q[sub]2[/sub] > 1 (as there are no integers with a difference smaller than 1). Then we have that (q[sub]1[/sub] - q[sub]2[/sub])b > b (I trust this is obvious). Take that in conjunction with our result that r[sub]2[/sub] - r[sub]1[/sub] < b, and you have a contradiction.

The proof for non-positive integers is similar enough that the reader should be able to supply it.

#10

He’s screwed up the positional system in the long division algorithm for one. More deeply, he’s assuming you can put any guess as to a partial quotient. I’ll explain 25/5 = 32:

First, he puts 3 instead of 5 as a number of times 5 goes into 25. Then he multiplies down to get 15, subtracts from 25 to get 10, forgets to drop down a zero for the next place, and proceeds with the division algorithm from there to get 2, which he just slaps next to the 3.

The important thing to note is that the integer division algorithm requires the largest number of times the divisor goes into the dividend without going over. Equivalently, in calculating a/b it states that there exist q and r>0 with |r|<|b| such that a=qb+r. This is uniquely defined (what he means to say by “absolute”).

#11

For his next trick, he’ll prove 28 / 7 = 13 . :rolleyes: :smiley:

#12

And 28 / 7 = 22
And 28 / 7 = 31

He’s misunderstood or misapplied the common algorithm for long division.

In the number systems consisting of fractions (rational numbers), real numbrs and complex numbers, division is unique, since every number x (except for 0) has a unique multiplicative inverse 1/x. And since multiplication is unique, we can define x/y as x * (1/y) (except, again, where y=0).

If you just take integers, then sometimes division is not defined, e.g., there is no integer equal to either 3/2 or 2/3

There are some algebraic systems where division is not unique, but in those systems, either:
(1) multiplication is not unique, or
(2) some numbers do not have a unique multiplicative inverse.

#13

By his own admission he has been shown to be wrong.
“NOTE: This was proven incorrect and I conceide with such. Read it for pure humour if you wish to. Or attempt to form your own theories out of it. I don’t care.” Quoted as the headiog of the web page linked to.
The poor guy doesn’t understand elementary principles of math. He’s in over his head and can’t understand the proof that he’s wrong.

#14

Not correct. We know that 28/7 = 13 because it was proven (not only by division, but by multiplication and addition) by the noted mathematicans L. Costello and B. Abbott.

#15

I bet this guy is totally baffled by the hotel problem, too. :rolleyes: