Understanding the given proof of integers - Ring theory

[chwala]

Homework Statement

see attached

Relevant Equations

Ring Theory

My interest is on the highlighted part ...

Now to my question,

in what cases do we have $mn<(m,n)[m,n]?$

I was able to use my example say,
Let $m=24$ and $n=30$ for example, then
$[m,n]=120$ and $(m,n)=6$ in this case we can verify that,
$720=6⋅120$ implying that, $mn≤ (m,n)[m,n]$.

 

[martinbn]

Now to my question,

in what cases do we have $mn<(m,n)[m,n]?$

This is a strange question, in the very next line the finish the proof, that it is an equality.

 

[chwala]

This is a strange question, in the very next line the finish the proof, that it is an equality.

I get your point the last line indicates an equal sign. However, ...the preceding line states that,
"Therefore, it must be less than the greatest common divisor'... on the contrary should it not be 'Therefore, it is equal to the greatest common divisor'? Unless there are cases where the inequality applies.

 

[fresh_42]

Homework Statement: see attached
Relevant Equations: Ring Theory

My interest is on the highlighted part ...

View attachment 330141

View attachment 330142

Now to my question,

in what cases do we have $mn<(m,n)[m,n]?$

Never. We have $\geq$ and $\leq$ making it $=$ and completing the proof.

I was able to use my example say,
Let $m=24$ and $n=30$ for example, then
$[m,n]=120$ and $(m,n)=6$ in this case we can verify that,
$720=6⋅120$ implying that, $mn≤ (m,n)[m,n]$.

The location with your red mark comes from $a\leq b \Longrightarrow a\cdot c\leq b\cdot c$ in case $c\geq 0.$ With $a=\dfrac{mn}{[m,n]}\, , \,b=(m,n)$ and $c=[m,n]$ we get what is written.

 

[chwala]

Never. We have $\geq$ and $\leq$ making it $=$ and completing the proof.The location with your red mark comes from $a\leq b \Longrightarrow a\cdot c\leq b\cdot c$ in case $c\geq 0.$ With $a=\dfrac{mn}{[m,n]}\, , \,b=(m,n)$ and $c=[m,n]$ we get what is written.

I can now see that two proofs that involve the inequalities $[≤]$ and $[≥]$ in general imply $[=]$, thus concluding the proof. Clear now...

 

[Mark44]

I can now see that two proofs that involve the inequalities $[≤]$ and $[≥]$ in general imply $[=]$, thus concluding the proof. Clear now...

This is a standard way of proving that two quantities are equal. If you can show that $a \le b$ and that $a \ge b$, then you can conclude that a = b.

 

[fresh_42]

I can now see that two proofs that involve the inequalities $[≤]$ and $[≥]$ in general imply $[=]$, thus concluding the proof. Clear now...

It is the same method that is usually used to show that two sets $A$ and $B$ are equal. We show $A\subseteq B$ ($a\in A \Longrightarrow a\in B$) and $B\subseteq A$ ($b\in B \Longrightarrow b\in A$) and conclude $A=B.$