My Math Forum Find a irrational number

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 March 18th, 2014, 01:58 AM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 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
 March 18th, 2014, 02:12 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 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: $\frac{\sqrt{n}}{99}$ Would do the trick. E.g. n = 37^2 + 1
 March 18th, 2014, 12:56 PM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Find a irrational number You can also do $\frac{38 \ - \ 37}{99} \= \ \frac{ 1}{99}$ $\frac{ 37}{99} \ + \ \frac{1}{99 \sqrt{n}} \= \ \frac{37 \sqrt{n} \ + \ 1}{99 \sqrt{n}} \ = \ \frac{37 n \ + \ \sqrt{n}}{99n}$ Will work for any NON SQUARE n > 1 Or similarly , $\frac{ 38}{99} \ - \ \frac{1}{99 \sqrt{n}}$ etc. , etc.

 Tags find, irrational, number

,
,

,

,

,

,

,

,

,

,

,

,

,

,

# find the iratinal number method

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post nfsmwbe Number Theory 13 April 17th, 2012 07:17 AM MyNameIsVu Number Theory 3 June 16th, 2009 08:13 PM habipermis Algebra 5 December 28th, 2008 02:54 PM jefferson_lc Real Analysis 2 September 5th, 2008 09:18 PM johnny Algebra 3 November 5th, 2007 03:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top