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.

## Discussions

Best Answer 9 years ago

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.

9 years ago

That is a trivial consequence of the

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.Answer 8 years ago

so 1/3 is irrational? but 1/2 isn't thats what you mean right?

Answer 8 years ago

No, 1/3 is rational because it repeats a pattern.

Answer 8 years ago

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

notrational precisely because youcannotfind any two integers with which you can write a ratio equal to pi.Answer 8 years ago

22/7 ?

Answer 8 years ago

They aren't equal. That's an approximation.

Answer 8 years ago

Im not that good at maths

Answer 9 years ago

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.

Answer 9 years ago

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.

8 years ago

the cherry filling

Answer 8 years ago

:)

8 years ago

math makes my head hurt

9 years ago

Answer 9 years ago

I would have voted that best answer had I seen it earlier.

9 years ago

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)

Answer 9 years ago

*is confused

Answer 9 years ago

about?

Answer 9 years ago

Math. It is really complicated for me. I think this is not so hard for you however...

Answer 9 years ago

it is actually kind of boring.

Answer 9 years ago

Can you explain what you just said a little more plainly?, or maybe elaborate on that equation?

Answer 9 years ago

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.

Answer 9 years ago

Nice, that explanation won something.

Answer 9 years ago

thank you.

Answer 9 years ago

THANK YOU! Now please post that as a new comment so that I can select it as a best answer.

9 years ago

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

simplepattern to pi.Answer 9 years ago

Just to clarify, it

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