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March 18th, 2014, 01:58 AM   #1
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Find a irrational number

Find an irrational number between 0.373737... and 0.383838...

I will use 0.(37) to mean 0.373737... and 0.(3 to mean 0.383838

My attempt
I take the smaller number 0.(37)
I can look at the zero but that will make the resulting number > 0.(3. This cannot be done. So the number will be 0."something"

I can look at the 3 but anything less than 3 will make the resulting number < 0.(37) and anything greater than 3 will make the resulting number > 0.(3. This cannot be done. So the number will be 0.3"something"

I can look at the 7. If I decrease it the resulting number < 0.(37) and if I increase it, it will equal or exceed 0.(3. This cannot be done. So the number is 0.37"something"

I can look at the next 3. I can increase it to any number (4 to 9) and it will be > 0.(37) but < 0.(3.

Possibilities are: 0.374..., 0.375..., 0.376..., 0.377..., 0.378..., 0.379...

I will arbitrarily take 0.374 and generate the irrational number 0.37401001000100001...
Am I correct?
Does this prove that there are infinite irrational numbers?
My reasoning goes like this
There's a fraction between any two fractions. Therefore there is an infinite number of fractions.
There's an irrational number between any two rational numbers. Therefore there is an infinite number of irrational numbers

Is there a better method of doing this kind of problem? Thanks
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March 18th, 2014, 02:12 AM   #2
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Re: Find a irrational number

I would calculate the two numbers you were given as limits of infinite geometric series:

0.373737... = 37/99 and 0.383838... = 38/99

Then, if n is any number between 37^2 and 38^2, then the square root of n will be irrational. So:



Would do the trick.

E.g. n = 37^2 + 1
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March 18th, 2014, 12:56 PM   #3
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Re: Find a irrational number

You can also do





Will work for any NON SQUARE n > 1

Or similarly ,



etc. , etc.

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