# If $x<1$ and $0.\overline{9}=1​$, then what is the maximum value of $x$?

How do you explain the maximum value of $x$ if $x<1$ and $0.\overline{9}=1​$? The mathematical notation is concise. But how can I explain this in words for a student in lower secondary school?

• the maximum value of x if x<1 There is no such maximum value, for example $(x+1)/2$ is always between $x$ and $1$. and 0.9¯=1​ That's a red herring, wholly unrelated to the previous question.
– dxiv
Commented Feb 24, 2017 at 2:47
• Why do you think that set bounded above should have a maximum? This is not guaranteed. What is guaranteed is that supremum of the set set exists under the same circumstances and if the supremum lies in the set then it becomes the maximum, otherwise no maximum exists (like the case here). Commented Feb 24, 2017 at 6:35
• … in words for a student in lower secondary school? We just finished looking at recurring decimals and then inequalities. I am trying to give them the language to discuss these concepts correctly. Commented Feb 27, 2017 at 6:11

You can say "There is no largest number that is less than $1$. In other words, if $x$ is any number less than $1$, there is another number bigger than $x$ that is still less than $1$."

The fact that $0.\overline{9}=1$ doesn't seem relevant to me.

• I wouldn't say it's irrelevent. If x < 1=.99999... them at least one digit is less than 9. But than there must be values larger that that by replacing that term with 9 but the further terms with numbers that aren't. I don't know that it helps or is a good explanation, but it isn't irrelevent. Commented Feb 24, 2017 at 3:32
• We just finished looking at recurring decimals and then inequalities, so it is relevant in the context of our current class discussions. But I will follow your advice and try to explain that “maximum value” is not a sensible thing to talk about when stating $x<1$. Commented Feb 27, 2017 at 6:13

If $x < 1=0.99999......$ then one of the decimal digits of $x$ must be something other than nine.

Say $x = 0.a_1a_2a_3......$ and $a_k \ne 9$. Then we can create $y =0.a_1a_2....a_{k-1}95$ and $x <y<0.a_1a_2...a_{k-1}99999....\le 0.99999.....=1$.

So for any $x <1$ there is always a $x <y <1$. So there can not ever be any maximum number less than $1$.

I don't know if that is the best argument, but it is an argument.