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I am trying to prove the density of rational numbers over the real line on my own.

I used the fact that any terminating number in base 10 is rational ( which I have proved already). So given any two real numbers a,b (no matter how close) I have proved One can construct a terminating number(similar to Cantor's proof of uncountability of irrational numbers) between the given numbers. Is this a valid proof? Or am I missing something?

jnyan
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  • somewhat related: http://math.stackexchange.com/questions/507899/proving-the-rationals-are-dense-in-r – Henricus V. Feb 19 '17 at 02:55
  • thank you But it seems to be standard proof of density of rational numbers, without any relation to representation of numbers in base 10. – jnyan Feb 19 '17 at 02:59
  • Seems valid although it also seems a bit convoluted. – Zaros Feb 19 '17 at 03:04
  • .This is a valid proof. If $r \in \mathbb R$ and $\epsilon >0$, then finding a terminating rational between $r$ and $r+\epsilon$ is an even stronger statement than density. Certainly, it implies density of rationals, moreover it implies density of terminating rationals. – Sarvesh Ravichandran Iyer Feb 19 '17 at 03:24
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    As @астонвіллаолофмэллбэрг said you have proven something even stronger than density. But there are some things to consider. Does your proof require your two numbers to have a decimal expansion? I have never seen a definition of the real numbers that uses decimal expansions since two expansions can represent the same number. Thus you may need to prove that there exists a decimal expansion for every real number (this has happened to me before) especially if this is within the scope of real analysis. If your proof does not require this then it should be fine. – Jeevan Devaranjan Feb 19 '17 at 03:48
  • @JeevanDevaranjan That's important, so thank you for pointing it out. – Sarvesh Ravichandran Iyer Feb 19 '17 at 03:49
  • ok. so if I prove that every real number has a decimal expansion ( I dont have to worry about uniqueness, as I just need atleast one representation) i will be done.?? – jnyan Feb 19 '17 at 03:54
  • @jnyan Yes then your proof would be sufficient. – Jeevan Devaranjan Feb 19 '17 at 09:52

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