# Is there a way to write 5.000000000 1?

1. Mar 26, 2003

### brum

ok so we can write greatest number that is < 5
(_
4.9 which means 4.9999999999999...

then.. what's the next highest number > 5?
` _
5.000000000000000...1 or 5.01

can you do that? (put the "repeating" there ^^)

edit: didn't realize the spaces would be removed so my repeating signs were at the beginning, not to where i spaced them out to

Last edited: Mar 28, 2003
2. Mar 26, 2003

### Integral

Staff Emeritus
4.999... = 5
Since there is no smallest number, it is not possible to represent the smallest number greater then something. Dealing with such concepts means you must specifiy intervals. The notation [1,2) implies all numbers greater then and equal to 1 but less then but not equal to 2. 1.999... = 2 so it is not included in the above interval, but any finite number of 9's is. Likewise (1,2] is the interval which contains all numbers greater then 1, but not 1, and less then or equal to 2. This is the concept of an open interval, that means the interval does NOT contain its endpoint.

5.000...1 where the elipsis represents an infinite number of 0s does not represent a real number. In that context it means a finite but unspecified number of 0s because by the definiton of real numbers the 1 MUST occupy a position which corresponds to some integer therefore there must be a finite number of 0s.

3. Mar 27, 2003

### steppenwolf

i believe this is known as a finite nonstandard number

4. Mar 28, 2003

### roeighty

maybe..

5 + 10^-n

=

5 + 1 / 10^n

(where n is natural number, that in this case gives arbitrary number of digits after the comma)
that, what you meant?

5. Mar 28, 2003

### Hurkyl

Staff Emeritus
There does not exist a greatest real number less than 5. ("decimal number" is synonymous with "real number")

There also does not exist a least real number greater than 5.

In fact, in any ordered number system with division and 2, "the greatest number less than x" never exists. Here is a proof by contradiction:

Suppose y is the greatest number less than x. This means that y < z < x can never be true for any z, because that would mean z is greater than y, but z is less than x.

Now, let z = (x + y) / 2

Starting from y < x
add y to both sides yielding y + y < x + y
this is the same as 2y < x + y
dividing by 2 yields y < (x + y) / 2
this is the same as y < z

Starting from y < x
add x to both sides yielding x + y < x + x
this is the same as x + y < 2x
dividing by 2 yields (x + y) / 2 < x
this is the same as z < x

so y < z < x... but this is impossible!

So our assumption that "y is the greatest number less than x" was false. Since there was no restriction placed on y (beyond being the greatest number less than x), there cannot exist a number that is the greatest number less than x.

As integral mentioned, 0.499... is, by definition, numerically equal to 0.5 (so, in particular, it cannot be less than 0.5)

Hurkyl

6. Apr 1, 2003

### vacuum

real numbers axiom

I suppose there should not be any number like that in existence according to an axiom which states that between any two reals there is another real number -> so between 5 and 5+eps (eps small as you want it to be) there is always another number.

Though, another question arises - is then 4.999.... same as 5 because 4.999... is larger than any other number smaller than 5?

7. Apr 1, 2003

### Integral

Staff Emeritus
Yes, As I posted above, 4.999... = 5. This can be demonstrated several different ways and can be rigoursly proven several different ways.