# Have I justified that $\forall x \in \mathbb{R}$, $x > 1 \rightarrow x^2 > x$

Have I justified that $\forall x \in \mathbb{R}$, $x > 1 \rightarrow x^2 > x$

Here is what I would do if this were asked on a test and I was told to "justify" the answer.

Let $x \in \mathbb{R}$

Assume $x$ is greater than $1$.

Then $x * x > x$ , since $x$ is greater than $1$.

Therefore $x^2 > x \square$

Not sure how that will fly with the grader or this community. What I would like to know is if I have correctly shown that the statement is true? How would you have justified that this is a true statement? It is these really obviously true or false statements that I have trouble proving.

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It depends on what level you are. If you were a first year's student and in certain courses focused on proving things, I wouldn't agree with the step that $x\cdot x >x$. However, in most other cases, I'd say you're doing too much. Be careful, though, with "greater than", since quite a lot of mathematicians tend to think that 1 is greater than 1 as well, since $1\geq 1$. (you'd want "strictly greater", or >, then) To decently prove it, I'd write $x=1+\epsilon$ for $\epsilon\in (0,\infty)$ and expand $x^2$. You'll be able to finish the prove then. –  HSN Jan 22 '13 at 0:35
@HSN I am not taking abstract linear algebra. So your first hypothesis is true. This course is an introduction to elementary discrete mathematics. –  Leonardo Jan 22 '13 at 0:37
That square I put there was a test to see how this community would accept it. My instructor claims that it is common to denote the end of a proof with this strange looking notation. –  Leonardo Jan 22 '13 at 0:40
@Leonardo, if by strange looking notation you mean the tombstone symbol, then yes. –  George V. Williams Jan 22 '13 at 1:03
I would specifically write the '$x*x \gt x$' bit of the proof as '$x*x \gt x*1$' and add a 'but $x*1=x$' line; it makes it much more clear what's going on. –  Steven Stadnicki Jan 22 '13 at 1:33

If you want it to more rigorous first prove that if $x \ge 0 \text{ and } a \ge b$ then $x*a \ge x*b$

The question that if you have proved it or not depends on what things you accept as obvious. For example in "Introduction to Set Theory" course this proof is not accepted.

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It indeed does fly. If you multiply both sides of an inequality by a positive quantity, the inequality is preserved.

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Note HSN's caveat: when you multiply by a quantity with an $x$ in it, be sure it is positive for all $x$! –  ncmathsadist Jan 22 '13 at 0:36

If $x > 1$, then $x > 0$ and $x-1 > 0$ so $x(x-1) > 0$ or $x^2-x > 0$ or $x^2 > x$.

An interesting variant on this is to show that if $x > 0$ then, unless $x = 1$, $x^2$ is further from $1$ than $x$ is and $\sqrt{x}$ is closer to $1$ than $x$ is. For extra credit, do this without separating into the cases $x < 1$ and $x > 1$.

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