#### Harte

##### Senior Member

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- 4,562

**Re: Second highest and lowest numbers**

An irrational number, like any other number, when expressed as a decimal has an infinite number of digits to the right of the decimal point. What makes the number irrational is that the digits to the right of the decimal point never repeat.

For example, the number three can be expressed as 3.000... with the zeroes repeating forever. The ratio one third (1/3) can be expressed .33333 with the threes repeating forever. But the number pi, as an irrational number, never has an infinitely repeating digit, or set of digits.

Now, we know pi is something ike 3.1415... on and on forever. We can ignore the 3 to the left of the decimal, which gives us another irrational number0.1415... This is a number between zero and one.

Now, I can't remember pi to 10 or twelve places, so I'm going to make these digits up, mainly just for the purposes of illustration. Lets call our irrational number;

0.141527658926...

Realizing that the digits in this number go on forever, it is easy to see that you could change one digit in this string and still have an irrational number, say change the 4 to a 3:

0.131527658926.

We now have a new irrational number, different from our first one.

We could take this new number and change one digit in it to generate another irrational number, and so on, and then repeat with our original number, only changing a different digit in the first step.

Also, let's take three consecutive digits and change them in an orderly fashion, say the digits 152:

0.131537658926

0.131637658926

0.132637658926.

Just continuing in this fashion gives us a thousand new irrational numbers. Anyone need any further evidence that the irrationals outnumber the rationals? Consider we are doing this by starting with just one irrational number. You could do it with the natural base e or the square root of two, or with any other rational number.

Also, start with a rational number that repeats. say 0.33333.

beginning with one of the 3's, add one to each three in a geometric progression (the second digit 3, the fourth digit 3, the eighth digit 3, and on forever) this turns 1/3 into an irrational number. Now do it again, only add 2 to each three in geometric progression. Now do it adding 3, now 4, ad infinitum. Now repeat the whole thing but use an exponential progression instead (add one to the 2nd, 4th, 16th, 32nd digits, etc.)

I think it's pretty plain now that there are more irrational numbers between zero and one than rational numbers on the entire number line.

Harte