Having problems in multiplying the inequalities

[Adesh]

Homework Statement

Can we multiply two inequalities element-wise?

Relevant Equations

$x \lt y$

Let’s say we are given two inequalities $$ x \lt y \\
a \lt b$$ then we can write (we can even prove it using logarithms) $$ ax \lt by$$ given that every number is positive.

In this article (Fact 4.1, point (ii) ) it is given that if $x, y$ are positive numbers and $a, b$ are negative numbers and if the following is the relation $$ x \lt y \\ a \lt b$$ then the multiplication would cause the inequality to flip over, I.e. $$ ax \gt b~y$$ .

But I have a counter example for that, $$ 2 \lt 3 \\ - 2 \lt -1 \\ -4 \lt -3$$ You see we got no flipping.

Please explain me.

 

[PeroK]

Homework Statement:: Can we multiply two inequalities element-wise?
Relevant Equations:: $x \lt y$

Let’s say we are given two inequalities $$ x \lt y \\
a \lt b$$ then we can write (we can even prove it using logarithms) $$ ax \lt by$$ given that every number is positive.

In this article (Fact 4.1, point (ii) ) it is given that if $x, y$ are positive numbers and $a, b$ are negative numbers and if the following is the relation $$ x \lt y \\ a \lt b$$ then the multiplication would cause the inequality to flip over, I.e. $$ ax \gt b~y$$ .

But I have a counter example for that, $$ 2 \lt 3 \\ - 2 \lt -1 \\ -4 \lt -3$$ You see we got no flipping.

Please explain me.

The article is clearly wrong, as your counterexample shows. Can you fix it?

 

[Adesh]

The article is clearly wrong, as your counterexample shows. Can you fix it?

No! I don’t have any access to it. It just came in the google search when I typed “multiplying the opposite inequalities”.

 

[PeroK]

No! I don’t have any access to it. It just came in the google search when I typed “multiplying the opposite inequalities”.

I didn't mean fix the document. I meant fix the inequality!

 

[Adesh]

I didn't mean fix the document. I meant fix the inequality!

Which inequality? I beg your pardon.

 

[PeroK]

Which inequality? I beg your pardon.

The one you posted about. 4.1 (ii). What should it be?

 

[Adesh]

The one you posted about. 4.1 (ii). What should it be?

Undetermined, as far as I can think ha?

 

[PeroK]

Undetermined, as far as I can think ha?

Maybe, maybe not.

 

[Adesh]

Maybe, maybe not.

Is there any rule which confirms that whether the inequality going to flip or not? Can we make sure whether it going to flip or not? Let’s consider $$ x\lt y $$ such that both of them are positive integers. And if $a,b$ are negative integers with the relation $$ a \lt b$$ then what conditions must be met for the following to be true $$ ax \lt b~y$$

 

[PeroK]

Let's focus on the case:

Let’s consider $$ x\lt y $$ such that both of them are positive integers. And if $a,b$ are negative integers with the relation $$ a \lt b$$

Although these need not be integers. What inequality must hold in that case?

 

[Adesh]

Let's focus on the case:
Although these need not be integers. What inequality must hold in that case?

$$-x \gt -y$$
$$- a \gt -b$$

 

[PeroK]

$$-x \gt -y$$
$$- a \gt -b$$

Okay, but we want something that involves all four numbers. I saw your attempted solution to that maths puzzle about the rationals. You were pretty close there. You're capable of working this out if you focus on it

 

[Adesh]

Okay, but we want something that involves all four numbers. I saw your attempted solution to that maths puzzle about the rationals. You were pretty close there. You're capable of working this out if you focus on it

Okay sir, I will post my thinking (about the inequality) here after a strong contemplation and I know you’re always there to help and encourage students.

 

[Adesh]

All right, $$ x \lt y ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x, y \in \mathbb {R^{+}} \\
a \lt b ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a, b \in \mathbb{R^{-}} \\
\implies |b| \lt |a| $$
So, we got two inequalities (involving positive numbers) going in the same side $$ x \lt y ~~~~~~~~~~(1) \\
|b| \lt |a| ~~~~~~~~(2)$$ Add (1) and (2) component wise $$ x +|b| \lt y + |a| ~~~~~~~~~~~~~~(3) $$
Now, subtracting $|a|$ from $x$ and $|b|$ from $y$ will obviously make the inequality even stronger, therefore we have $$ x - |a| \lt y - |b|~~~~~~~~~~~~~~~~~~~(4) $$

Multiply inequality $(3)$ and $(4)$ component -wise, $$ ( x + |b|) ~(x - |a|) \lt (y+ |a|) ~(y-|b|) \\
x^2 - |a|x + |b|x - |a||b| \lt y^2 - |b|y + |a|y - |a||b| \\
x^2 -|a|x + |b|x \lt y^2 -|b|y +|a|y$$
Since, it is very obvious that if $x \lt y$ then $x^2 \lt y^2$, therefore the above inequality may be written as $$ |b|x - |a|x \lt |a|y - |b|y ~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)$$

From (1) and (2) we have $|b|x \lt |a| y$ (as each and every number is positive) and therefore $0 \lt |a|y - |b| x$. Writing out inequality (5) in this fashion
$$ |b| y - |a| x \lt |a|y - |b|x $$ Although, out RHS of the above inequality is positive but since LHS is less than our RHS, hence no judgement can be done about LHS. It may be positive or negative. That is we have two cases without any determination about which of them is true $$ |b|y - |a|x \lt 0~~~~~~~~~~~~~~~~~~~~~~(i) \\
\mathbf{OR} \\

|b|y - |a|x \gt 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$$

Doing some little algebra with the above inequalities $$ |b|y \lt |a|x \\
ax \lt b~y ~~~~~~~~~~~~~~~~~~~(\textrm{to remove the modulus we multiplied both sides by negative 1}) $$
$$ \mathbf{AND} \\

|b|y \gt |a|x \\
b~y \lt ax $$ That is, it is not possible to conclude whether the inequality going to flip or not.

 

[PeroK]

The pdf had a simple typo. It should have been:
$$0 < A < B\ \text{and} \ C \le D < 0 \ \Rightarrow \ BC < AD < 0$$

 

[Adesh]

The pdf had a simple typo. It should have been:

0<A<B and C≤D<0 ⇒ BC<AD<00<A<B and C≤D<0 ⇒ BC<AD<0​

Yes, wow!

$$A \lt B \\ |D| \lt |C| \\ A|D| \lt B|C| \\ AD \gt BC $$

 

[PeroK]

Yes, wow!

That shouldn't have been hard to work out.

 

[Adesh]

That shouldn't have been hard to work out.

I was just writing it :)

 

[Adesh]

That shouldn't have been hard to work out.

Have I understood the things the way you wanted me, sir?

 

[PeroK]

Have I understood the things the way you wanted me, sir?

You don't need to call me sir! Yes, it was just to work out that there is an inequality, but not the one in the article.