# Equality of real numbers : transitivity holds or not with this definition ??

I was reading this book called "The Taylor series" by Paul Dienes. He wrote and I quote, "We say that the positive real number $a=a_0.a_1a_2a_3a_4...$ Precedes $b=b_0.b_1b_2b_3..$ if $a_0<b_0$ or $a_0=b_0 ,a_1=b_1, a_2=b_2.... a_{n-1} =b_{n-1}$ But $a_n<b_n$ . We say that b is between a and c if either a precedes b and b precedes c OR c precedes b and b precedes a. Definition of equality :- a=b if there is no number between a and b. " Now my question is what happens to transitivity of equality in this case? If there is no number between a and b a=b . If there is no number between b and c ,b=c. And transitivity now implies a=c , but how can it be since b lies between a and c?? Also, suppose a=1.9999.... and b =2 , then we know a

• Where did you get that $b$ is between $a$ and $c$ from? – Arthur Sep 14 '18 at 11:36
• @Arthur Also, suppose a=1.9999.... and b =2 , then we know a <b . So a definitely precedes b. And by the definition of equality a=b. Can these 2 hold together?? – Kavita Juneja Sep 14 '18 at 13:47

You have to convince yourself that the only way how string $a$ can precede string $b$ with no string in between is when $$a=a_0.a_1a_2\ldots a_{r-1} a_r99999\ldots,\qquad b=a_0.a_1a_2\ldots a_{r-1} b_r00000\ldots\ ,$$ whereby $0\leq a_r<9$ and $b_r=a_r+1$. It then follows that the situation you are afraid of cannot occur.