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.

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

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.

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.

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.

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.

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)

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.

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.

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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.

definitionof irrational. Arational numberis one which is expressed in the form of a ratio of integers (a fraction). The numerator and denominator may be arbitrarity large, such as12345678901234567890 / 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.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

notrational precisely because youcannotfind any two integers with which you can write a ratio equal to pi.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)

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.

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

simplepattern to pi.ispossible to actually any desired digit of pi, it is the statistics of the answers that have have yet to be proven.