0.9_ = 1?

Post Reply New Topic

I think it's conceivable that robo37 isn't trolling. If you told me about 0.9_=1 when I was fourteen I'd probably be quite puzzled.

Anyway, your definition of an integer as a real number with everything right of the decimal point being zeroes fails, and it fails precisely because you can express 1 as 0.9_. Instead the set of integers is considered (explicitly) as the set of the natural numbers with zero and the negatives of the natural numbers.

Last edited by Hector on 26 Jan 2009, 3:28 pm, edited 2 times in total.

I call it a waste of time, since there is yet no practical need for me to divide anything by zero. My son, the maths major, can wax eloquent for hours on the various theoretical uses for 1/0, but he has yet to show me how I could use it in my work.

This is slightly off topic, but when your dividing by zero your basically asking what do you times zero by to get X. If X is a non-zero number then the question has no answer because even infinity multiplied by zero is zero. If X is zero then the answer can be any number, or X if you prefer. You can see this clearer if you look at this;

Y*X=YX
YX/Y=X

so if Y=0...

0*X=0
0/0=X

It is impossible to have X/0 through this.

Last edited by robo37 on 26 Jan 2009, 4:12 pm, edited 1 time in total.

All integers are real numbers. But not all real numbers are integers or even rational numbers (numbers that can be expressed as fractions).

ruveyn

Illustrating that they are the same isn't too bad. From introductory set theory we find out that between any two real numbers is another real number. If .9_ and 1 are different numbers, then there is a number between them. The problem as the reader is probably aware, is that the only way to find a number GT than .9_ would be to have a that number be GT 9 at some decimal place in .9_. Unfortunately this also results in a number that is GE 1, so we find by contradiction that .9_ must be the same as 1.

Understood, but if you were correcting me I'm not sure exactly what you were addressing.

In what's called a "division ring", division by a number is understood as multiplication by that number's inverse. Thus you'd need 0*(0^-1) = 1. The problem with that is that it is an immediate consequence of the axioms of any ring that 0*anything = 0. Thus the introduction of division by zero destroys the ring's structure and the algebra.


Quite the case. Most of the confusion people have is that they don't recognize the inherently limit based definition of repeating decimals.

So, by that rule, 9 must be 10 ? or 1 must be 0

What? ಠ__ಠ

I know a few engineers who brought up the question, they call it infinity and argue that mathematicians shouldn't call it undefined. And I pointed out (such as the case in this thread) that since x never is actually defined at 0 it is better to say undefined, instead of using the limit definition of infinity- without specifying that you are using a limit definition and not overlooking the behavior of division by 0. Though technically x does in fact go into numbers an infinite time during division.

you could probably disprove the .999 repeating decimal is 1 assumption in any number of mathematical ways starting with set theory and the fact that 1 is a natural number and an integer, while .9999... is not. 1 is rational and .999 is not. Identity properties of 1, etc.

The only people I've ever seen use 1/0 = infinity certainly haven't been mathematicians. Physicists and economists have been the worst offenders in my studies.

Let's see you try.

Last edited by twoshots on 27 Jan 2009, 10:16 am, edited 2 times in total.

Wrong. One cannot disprove what has been correctly proven.

ruveyn

If you ask someone what is the closest number to 1 that’s not 1 then they probably say 0.9_, but if you tell them that 0.9_ is 1 then they would probably say 1.0_1, but if you then tell them that that isn't even a number they would just guess and say something like 0.9, which obviously isn't the closest number because 0.99 is closer, but 0.999 is closer than 0.99 and 0.9999 is closer than 0.999 ect. This means that the closest number to one is zero with a infinite number of nines after the decimal point, which means the closest number to 1 is 0.9_ so 0.9_ cant be the same as 1. 1-0.9=0.1, 1-0.99=0.01, 1-0.999=0.001, 1-0.9999=0.0001, 1-0.99999=0.00001... so one minus zero with an X number of nines after its decimal point equals a number that is greater than zero. This would apply even if X is infinity, in the same way that 1/0 isn’t infinity because 0*infinity is still 0.

You ignore the alternative which is true: there is no "closest" number to 1. This is because both the real and rational numbers are "dense": between any two numbers, there is another one. The sequence (.9,.99,.999,...) which you have correctly identified with .999... approaches 1 as the number of 9's becomes infinite, so really you've just shown that if you tried to find a number closest to 1 by that method you can't and instead wind up at 1 itself.

you have correctly proven that .9 repeating is 1?


lolz