Prove by contradiction that a real number that is less than every positive real number cannot be posisitve This is an question from the book "A concise introduction to Pure Mathematics". I understand that it looks like a homework question but it's the first chapter and there are no answers for even questions. As I am independently trying to make my way through a bit of maths I was hoping I could get some help.
The real reason I am stumped is I don't yet have a firm grasp on what a proof is (at which point its conclusively proven).
I think the negation of the statement would be
"There exists a real number less than every positive real number that is positive"
I am tempted to use square root and say
n > 0
n = sqrt(k)
therefore k > 0
I don't think I have proven anything.
Also: Could someone please advise and perhaps throw some beginner texts my way?
Thank you
 A: I would like to point out that the answer to your question depends very much on the exact phrasing. I very much doubt that the question you posted is 1:1 the question in the book.
In some regions of the world, cough e.g. Germany as well as France, zero is considered to be both positive and negative at the same time, see for instance here: Is zero positive or negative?
In fact, if zero is considered positive, then there exists a real number that is smaller than every other positive real number and still positive, namely zero.
Why I'm pointing this out is the following: Were the question states as you posted it here, i.e. "Prove by contradiction that a real number that is less than every positive real number cannot be positive."
Then the proof is simple, because
$$
\forall x \in \mathbb{R}: \neg (x \lt x)
$$
So the number in your question would have to be smaller than itself in order to fulfill the stated requirement anyway, which clearly does not hold.
A: Suppose that $x$ is a real number less than every positive number. Suppose to the contrary that $x$ is positive. Then $x>x$.
A: Perhaps easier: if $\;\epsilon>0\;$ is less than any positive number, then also
$$\epsilon<\frac\epsilon2\;,\;\;\text{since also}\;\;\frac\epsilon2>0\;\;\;\ldots\;\;\text{contradiction}$$
