2

What am I doing wrong?

\documentclass{amsart}
\usepackage{amssymb,amsmath,latexsym,times,color}


\newtheorem*{theorem*}{Theorem}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\begin{document}

Question Number 3
\begin{theorem*}
Since a and b are positive numbers, then $$\frac{a}{b}$$ and $$\frac{a+2b}{a}$$ are positive integers.
\end{theorem*}
\begin{proof}
Case 1: If $$\frac{a}{b}\leq 2$$ then $$min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq\frac{a}{b}\leq 2$$
Thus $$min{\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq 2$$
Case 1: If $$\frac{a}{b}\leq 2$$ then $$min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq\frac{a}{b}\leq 2$$
Thus min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq 2
Case 2: If $$\frac{a}{b}>2$$ then $$\frac{b}{a}<(1/2)$$
Since $$\frac{a+2b}{a}= \frac{a}{a} + \frac{2b}{a}$$ which also means it is < 1+2*(1/2)
Thus $$\frac{a+2b}{a}<2$$
so $$min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq\frac{a+2b}{a}<2$$
Hence from both sides we have min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq 2
If you do this again with opposite inequalities you will get:
Case 1: If $$\frac{a}{b}\leq2$$ then \frac{b}{a}\geq (1/2)
Since $$\frac{a+2b}{a}= 1+\frac{2b}{a}\geq 1+2*(1/2)= 2$$ so \frac{a+2b}{a}\geq 2
thus $$max{\frac{a}{b},\frac{a+2b}{a}}\geq \frac{a+2b}{a}\geq 2$$ which simplifies to:
$$max{\frac{a}{b},\frac{a+2b}{a}}\geq 2$$
Case 2: If $$\frac{a}{b}>2$$ then $$max{\frac{a}{b},\frac{a+2b}{a}}\geq \frac{a}{b}>2$$
This means:
$$max{\frac{a}{b},\frac{a+2b}{a}}>2$$ 
From both cases we get: $$max{\frac{a}{b},\frac{a+2b}{a}}\geq 2$$
Since both sets of inequalities are true, we obtain:
$$min{\frac{a}{b},\frac{a+2b}{a}}\leq 2\leq max{\frac{a}{b},\frac{a+2b}{a}}$$
\end{proof}
\end{document}
4

The following is a working example, I'm sure some mathematician here would improve it.

If it doesn't work on your computer, you have some problems with your TeX distribution.

I highly recommend to you to read What are good learning resources for a LaTeX beginner?

\documentclass{amsart}
\usepackage{amssymb,amsmath,newtxtext,newtxmath}

\newtheorem*{theorem*}{Theorem}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}

\begin{document}
Question Number 3
\begin{theorem*}
    Since a and b are positive numbers, then 
    \[
    \frac{a}{b}
    \] 
    and 
    \[
    \frac{a+2b}{a}
    \] 
    are positive integers.
\end{theorem*}
\begin{proof}
    Case 1: If 
    \[
    \frac{a}{b}\leq 2
    \] 
    then 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq\frac{a}{b}\leq 2
    \]
    Thus 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq 2
    \]
    Case 1: If 
    \[
    \frac{a}{b}\leq 2
    \] 
    then 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq\frac{a}{b}\leq 2
    \]
    Thus 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq 2
    \]
    Case 2: If 
    \[
    \frac{a}{b}>2
    \] 
    then 
    \[
    \frac{b}{a}<(1/2)
    \]
    Since 
    \[
    \frac{a+2b}{a}= \frac{a}{a} + \frac{2b}{a}
    \] 
    which also means it is \(< 1+2*(1/2)\).
    Thus 
    \[
    \frac{a+2b}{a}<2
    \]
    so 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq\frac{a+2b}{a}<2
    \]
    Hence from both sides we have 
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq 2
    \]
    If you do this again with opposite inequalities you will get:
    Case 1: If 
    \[
    \frac{a}{b}\leq2
    \] 
    then 
    \[
    \frac{b}{a}\geq (1/2)
    \] 
    Since 
    \[
    \frac{a+2b}{a}= 1+\frac{2b}{a}\geq 1+2*(1/2)= 2
    \]
    so 
    \[
    \frac{a+2b}{a}\geq 2
    \] 
    thus 
    \[
    \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\geq \frac{a+2b}{a}\geq 2
    \] 
    which simplifies to:
    \[
    \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\geq 2
    \]
    Case 2: If 
    \[
    \frac{a}{b}>2
    \]
    then 
    \[
    \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\geq \frac{a}{b}>2
    \]
    This means:
    \[
    \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}>2 
    \]
    From both cases we get: 
    \[
    \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\geq 2
    \]
    Since both sets of inequalities are true, we obtain:
    \[
    \min\left\{\frac{a}{b},\frac{a+2b}{a}\right\}\leq 2\leq \max\left\{\frac{a}{b},\frac{a+2b}{a}\right\}
    \]
    \end{proof}
\end{document}

enter image description here enter image description here

  • thank you! it doesn't work on my computer but i used an online editor to get the pdf version of my work in progress. but this is super helpful. – Britney Sep 11 '17 at 6:05
  • 2
    @Britney I would guess you are using the incorrect command on your local installation (tex rather than latex) – David Carlisle Sep 11 '17 at 6:36
2

Code such as

Thus min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq 2
...
so $$min{$$\frac{a}{b}$$,$$\frac{a+2b}{a}$$}\leq\frac{a+2b}{a}<2$$

is simply incorrect. I suspect that what you really want is

Thus $\min(\frac{a}{b},\frac{a+2b}{a})\leq 2$
...
so $\min(\frac{a}{b},\frac{a+2b}{a})\leq\frac{a+2b}{a}<2$

Note that $...$ serves to enter and exit inline math, whereas $$...$$ serves to enter and exit display-style math. In fact, $$...$$ shouldn't even be used in a LaTeX document; see the posting Why is \[ … \] preferable to $$ … $$? for more information on this particular subject.

The times package is deprecated; I suggest you load the newtxtext and newtxmath packages.

\min and \max do not take arguments. If you want to delimit their scope, use (...), [...], or \{...\} -- not {...}.

Assuming that you mostly want inline math in the proof environment, except maybe for the very last equation, the following may be of use to you. (By the way, I haven't checked the math itself!!)

enter image description here

\documentclass{amsart}
\usepackage{amssymb,amsmath,newtxtext,newtxmath}

\newtheorem*{theorem*}{Theorem}
\begin{document}

\begin{theorem*}
If $a$ and $b$ are positive numbers, then $\frac{a}{b}$ and $\frac{a+2b}{a}$ are positive integers.
\end{theorem*}
\begin{proof}
Case 1: If $\frac{a}{b}\leq 2$ then $\min\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\leq\frac{a}{b}\leq 2$.
Thus $\min\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\leq 2$.

Case 2: If $\frac{a}{b}>2$ then $\frac{b}{a}<(1/2)$.
Since $\frac{a+2b}{a}= \frac{a}{a} + \frac{2b}{a}$ which also means it is less than $1+2(1/2)$.
Thus $\frac{a+2b}{a}<2$
so $\min\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\leq\frac{a+2b}{a}<2$.

Hence from both sides we have $\min\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\leq 2$.

If you do this again with opposite inequalities you will get:

Case 1: If $\frac{a}{b}\leq2$ then $\frac{b}{a}\geq (1/2)$.
Since $\frac{a+2b}{a}= 1+\frac{2b}{a}\geq 1+2(1/2)= 2$ so $\frac{a+2b}{a}\geq 2$
thus $\max\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\geq \frac{a+2b}{a}\geq 2$ which simplifies to
$\max\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\geq 2$

Case 2: If $\frac{a}{b}>2$ then $\max\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\geq \frac{a}{b}>2$.
This means:
$\max\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)>2$.

From both cases we get: $\max\bigl(\frac{a}{b},\frac{a+2b}{a}\bigr)\geq 2$.

Since both sets of inequalities are true, we obtain:
\[
\min\Bigl(\frac{a}{b},\frac{a+2b}{a}\Bigr)\leq 2\leq 
\max\Bigl(\frac{a}{b},\frac{a+2b}{a}\Bigr) \qedhere
\]
\end{proof}
\end{document}

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.