# What causes irrational numbers such as pi to never form patterns?

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geodez says: May 19, 2009. 8:38 PM
here it is:

Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.

It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2.

This means b2 is even, from which follows again that b itself is an even number!!!
WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.
kelseymh says: Apr 27, 2009. 12:35 PM
That is a trivial consequence of the definition of irrational. A rational number is one which is expressed in the form of a ratio of integers (a fraction). The numerator and denominator may be arbitrarity large, such as

12345678901234567890 / 9999998888888777777766666665555553

but they are both still integers. All rational numbers, when expressed in the form of a "decimal" (in any base, not just base ten) will either terminate with a finite number of digits (followed by zeros :-), or will form a repeating pattern.

The converse is also true. Any decimal expression which forms a terminating or repeating pattern can be expressed as a fraction

All numbers which cannot be expressed as fractions, and which therefore must have an infinite series of non-repeating decimal digits, is called irrational.
knex_mepalm in reply to kelseymhOct 15, 2009. 9:18 PM
so 1/3 is irrational? but 1/2 isn't thats what you mean right?
Torpe in reply to knex_mepalmSep 15, 2010. 3:10 PM
No, 1/3 is rational because it repeats a pattern.
kelseymh in reply to knex_mepalmOct 16, 2009. 8:33 AM
No, 1/3 is rational -- trivially, as you just wrote it as a ratio (that's where the word "rational" comes from) of two whole numbers (1 and 3).  Did you actually read what I wrote before replying to it?

Pi is not rational precisely because you cannot find any two integers with which you can write a ratio equal to pi.
darkpid in reply to kelseymhJun 22, 2010. 3:46 PM
22/7 ?
kelseymh in reply to darkpidJun 22, 2010. 4:25 PM
They aren't equal. That's an approximation.
knex_mepalm in reply to kelseymhOct 16, 2009. 10:32 PM
Im not that good at maths
NobodyInParticular in reply to kelseymhApr 27, 2009. 1:49 PM
Oblivitus didn't specify whether the 'pattern' had to be repeating, so I went with answering 'Why is pi normal?' rather than 'Why is pi irrational?'. I may have been reading too much into it, though.
kelseymh in reply to NobodyInParticularApr 27, 2009. 1:57 PM
I guess I assumed that someone asking the more technical question must know that definition of "normal," and would likely have used the technical term rather than a vague lay-person's description. Hence my more simpleminded explanation. The issue of normality is fairly subtle and theoretical.
SnapFitPieces(: says: Mar 4, 2010. 11:22 AM
the cherry filling
Oblivitus (author) in reply to SnapFitPieces(:Mar 4, 2010. 6:13 PM
:)
DELETED_Haon says: Feb 26, 2010. 6:56 PM
NachoMahma says: Apr 27, 2009. 10:36 AM
. It just does. heehee

Oompa-Loompa in reply to NachoMahmaJul 3, 2009. 8:08 PM
I would have voted that best answer had I seen it earlier.
geodez says: May 19, 2009. 11:11 AM
Pi is transcendental, which means that
it is not a root of any polynomial equation with rational
coefficients. (Note that every rational number P/Q, with P and Q
integers, is a root of QX-P = 0)
TigerNod in reply to geodezJun 20, 2009. 7:53 AM
*is confused
geodez in reply to TigerNodJun 20, 2009. 8:07 AM
TigerNod in reply to geodezJun 22, 2009. 11:45 AM
Math. It is really complicated for me. I think this is not so hard for you however...
geodez in reply to TigerNodJun 22, 2009. 3:52 PM
it is actually kind of boring.
Oblivitus (author) in reply to geodezMay 19, 2009. 1:41 PM
Can you explain what you just said a little more plainly?, or maybe elaborate on that equation?
geodez in reply to OblivitusMay 19, 2009. 3:06 PM
this is going to take a while. i will use the square root of 2 as an example.
Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally make it so that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction.

It follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2.

This means b2 is even, from which follows again that b itself is an even number!!!
WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.
Oblivitus (author) in reply to geodezMay 20, 2009. 7:01 PM
Nice, that explanation won something.
geodez in reply to OblivitusMay 21, 2009. 7:15 AM
thank you.
Oblivitus (author) in reply to geodezMay 19, 2009. 6:04 PM
THANK YOU! Now please post that as a new comment so that I can select it as a best answer.
NobodyInParticular says: Apr 27, 2009. 12:44 PM
It has been proved that pi is irrational (it cannot be written as a fraction of two whole numbers) and that it is trancendental (it cannot be written as a polynomial with rational coefficients). For highschool-level math classes, this is generally sufficient to assume that the digits are 'as good as random'. If you are interested in the proofs, a little calculus is needed.

However, whether the digits appear with equal frequency is a very tricky question that has yet to be solved definitively.

Still, given how many digits of pi have been calculated so far, I think it is fairly safe to say that we will not be finding any simple pattern to pi.
NobodyInParticular in reply to NobodyInParticularApr 27, 2009. 1:45 PM
Just to clarify, it is possible to actually any desired digit of pi, it is the statistics of the answers that have have yet to be proven.