1 = 1

[ak00ma]

if n is equal to .9 and you multiply n with 10 then it equals 9 and not 9.9, or did I missunderstand something???

[SCM]

n = 0.99999999999999999999999999999999999999999999...
It keeps repeating forever. The formal mathematical method for proving it involves showing that you can't find any number between 0.9... and one. If no number exists between two numbers then they are really the same number.

[oracle]

n = 0.99999999999999999999999999999999999999999999...
It keeps repeating forever. The formal mathematical method for proving it involves showing that you can't find any number between 0.9... and one. If no number exists between two numbers then they are really the same number.

Correct. I think it is called "dense" when there is no number between 2 numbers.

So, anyone know how to prove that root 2 aint a fraction?

[Flexibal]

yes, tis simple:

let the square root of 2 be a rational number, which we'll represent as a/b.

so

sqr(2) = a/b
2 = a^2 / b^2
2 * b^2 = a^2

now, according to the rules of even numbers, any number multiplied by two is even. so (2*b^2) is even, and since (a) equals it, (a^2) is even as well.

another rule of even numbers is that an even number squared is an even number. since we already found out that (a^2) is an even number, it means (a) is an even number.

because a is even, we can represent it an a multiple of 2 and a natural number:

a = 2*c

we get

2*b^2 = (2*c)^2
2*b^2 = 4*c^2
b^2 = 2*c^2

hence (b^2) is also even, and therefore (b) is even as well.

now, if (a) and (b) are even numbers, they are both multiples of 2, which means we can reduce them into a simpler form.

a = 2*c
b = 2*d
a/b = (2*c)/(2*d) = c/d

so

a/b = c/d = sqr(2)

and this whole process will start over - we'll see that both © and (d) are even numbers.... etc. this means that the fraction that represents sqr(2) can be always reduced by 2, which contradicts one of the rules of rational numbers (can be reduced only a finite number of times).


therefore, sqr(2) cannot be represented as a rational number.

---

also, i came up with some hypothesis of my own.

now rational number squared can result in an integer, except for integers. e.g.

integer^2 = integer

rational-that-is-not-an-integer ^ 2 != will-never-be-an-integer

irrational ^ 2 = may-be-an-integer

like the square root of 2 is irrational, but squaring it results in an rational number (2)... while (a/b)^2, where a, b are integers, a!=b, and b!=1, will never result in an integer.

this hypothesis is quite interesting, because it will allow you to prove that any square root that is not an integer, is necessarily irrational.

so we can prove that sqr(2) is irrational, but how could you prove that sqr(3) is irrational just as well? that even-numbers trick will not work here....

so if my hypothesis is correct, you can say that any non-integer square root is irrational.

----




[Flexibal>

[Agamemnus]

3 + 3 = 6
6 - 3 = 3
3 * 3 = 9
9 / 3 = 3
3 + 3 - 3 * 3 / 3 = 3

is:

(a + a - a) * a / a

which is:

a + a - a

which is:

a.

Erroneously assuming it is:

"a + a - a * a / a = a + a - a = a" gives the same answer. But it's WRONG!

[SCM]

Agamemnus,

Try

Code:

INPUT "Enter a number";a
PRINT "a ="; a
PRINT "(a + a - a) * a / a ="; (a + a - a) * a / a
PRINT "a + a - a * a / a ="; a + a - a * a / a

It is a true statement even though your reasoning was not very careful. Again according to order of operations, work should be done inside parentheses before multiplying (or you should distribute). Since the multiplied term, a / a, is equal to 1, the multiplicative identity, doing it the way that you did it did not change anything.

Looking at your original expression:

Code:

(a + a - a) * a / a
=(a + (a - a)) * (a / a)           by the associative properties of addition and multiplication
=(a + 0) * 1                       by additive and multiplicative inverses
= a * 1                            by the additive identity
= a                                by the multiplicative identity

And starting with the left side of the final equation:

Code:

a + a - a * a / a
= a + a - a * (a / a)               by the associative property of multiplication
= a + a - a * 1                     by multiplicative inverses
= a + a - a                         by the multiplicative identity
= a + (a - a)                       by the associative property of addition
= a + 0                             by additive inverses
= a                                 by the additive identity

What was it that you were saying was wrong?

Flexibal,

The square root of some real numbers is rational. For example SQR(.25) = .5. The square root of any prime number can be shown to be irrational in a proof similar to yours, and taken further the square root of any integer that is a not a perfect square is irrational.

[Flexibal]

foxy, i meant the square root of integral numbers. the number theory usually researches the properties of integral numbers, not fractional ones.


and i think i can prove my claim. let (a) and (b) be positive integral numbers, having no common factors except for 1, and b != 1. let © be a positive integral number as well, let (a/b) be the square root of ©.

let [a1, a2, a3, ... an] be the prime factors of (a), and [b1, b2, b3, ..., bn] be the prime factors of (b).

sqr© = a/b = (a1*a2*a3*...*an) / (b1*b2*b3*....*bn)
c = (a/b)^2 = (a1*a2*a3*...*an)^2 / (b1*b2*b3*....*bn)^2 = (a1*a2*a3*...*an * a1*a2*a3*...*an) / (b1*b2*b3*....*bn * b1*b2*b3*....*bn)

since (a) and (b) have no common factors, (a/b)^2 will never be whole. since © is whole, and (a/b)^2 = c, it means that no rational number (expect for integers) can solve that.

therefore, we resort to irrational numbers. so any non-whole square root of an integer is necessarily irrational.

therefore, sqr(2), sqr(3), sqr(7), .... are all irrational and you do not need to prove it like the proof i gave for sqr(2).









[Flexibal>

[Agamemnus]

I said the same thing. The result is the same, but the method which some previously used, without recognizing the order of the operations, is wrong....