Proving inequality properties with case analysis

[Kernul]

Homework Statement

Prove the following facts about inequalities. [In each problem you will have to consider several
cases separately, e.g. $a > 0$ and $a = 0$.]
(a) If $a \leq b$, then $a + c \leq b + c$.
(b) If $a \geq b$, then $a + c \geq b + c$.
(c) If $a \leq b$ and $c \geq 0$, then $ac \leq bc$.
(d) If $a \leq b$ and $c \leq 0$, then $ac \geq bc$.

Homework Equations

The Attempt at a Solution

So, I've tried to prove the first one (the second is basically the first one but with inequalities inverted) the following way:
If $a > 0, b > 0$ or $a < 0, b < 0$ (do I really have to say that $b> 0$ too?)
$a \leq b \implies a + c \leq b + c$
$a + (c + (-c)) \leq b + (c + (-c))$
$(a + c) + (-c) \leq (b + c) + (-c)$
$a + c \leq b + c$
If $a = 0, b > 0$
$0 \leq b \implies c \leq b + c$
$(c + (-c)) \leq b + (c + (-c))$
$c + (-c) \leq (b + c) + (-c)$
$c \leq b + c$
So it's proved, right?
Now, going to (c), we have this $c \geq 0$ that stops me from using $c^{-1}$ to prove them because it's not $c > 0$.
Which other way can I prove the last two?

EDIT: Sorry, I didn't notice the inequalities were wrong.

 

[Ray Vickson]

Homework Statement

Prove the following facts about inequalities. [In each problem you will have to consider several
cases separately, e.g. $a > 0$ and $a = 0$.]
(a) If $a <= b$, then $a + c <= b + c$.
(b) If $a >= b$, then $a + c >= b + c$.
(c) If $a <= b$ and $c >= 0$, then $ac <= bc$.
(d) If $a <= b$ and $c <= 0$, then $ac >= bc$.

Homework Equations

The Attempt at a Solution

So, I've tried to prove the first one (the second is basically the first one but with inequalities inverted) the following way:
If $a > 0, b > 0$ or $a < 0, b < 0$ (do I really have to say that $b> 0$ too?)
$a <= b \implies a + c <= b + c$
$a + (c + (-c)) <= b + (c + (-c))$
$(a + c) + (-c) <= (b + c) + (-c)$
$a + c <= b + c$
If $a = 0, b > 0$
$0 <= b \implies c <= b + c$
$(c + (-c)) <= b + (c + (-c))$
$c + (-c) <= (b + c) + (-c)$
$c <= b + c$
So it's proved, right?
Now, going to (c), we have this $c >= 0$ that stops me from using $c^{-1}$ to prove them because it's not $c > 0$.
Which other way can I prove the last two?

Looking at several cases for each part is a waste of time; in every part there is no need at all for "cases". Just use an appropriate definition of "≤" or "≥".

 

[Kernul]

Looking at several cases for each part is a waste of time; in every part there is no need at all for "cases". Just use an appropriate definition of "≤" or "≥".

An appropriate definition of $\leq$ and $\geq$? Wouldn't that be:
$a \leq b$ means $a < b$ or $a = b$
$a \geq b$ means $a > b$ or $a = b$

 

[Ray Vickson]

An appropriate definition of $\leq$ and $\geq$? Wouldn't that be:
$a \leq b$ means $a < b$ or $a = b$
$a \geq b$ means $a > b$ or $a = b$

What does $a < b$ mean? What is a very simple way to test if $a < b$?

Anyway, I did not claim that what you are doing is wrong; I am just saying it is not the fastest way.

To deal with your original question: in your argument you started with $a \leq b$, then got $(a+c) + (-c) \leq (b+c) + (-c)$, then canceled the "$(-c)$' on both sides to end up with $a+c \leq b+c$. However, that last step is invalid: it basically assumes what you are trying to prove, because if you set $A = a+c$ and $B = b+c$ you are going from $A -c \leq B-c$ to $A \leq B$---but you have not proved that yet.

 

[Kernul]

What does $a < b$ mean? What is a very simple way to test if $a < b$?

$a < b$ means that $a$ is a number smaller than $b$.
To see if $b - a$ is still a positive number? If it's negative it means that actually $a > b$.

 

[Ray Vickson]

$a < b$ means that $a$ is a number smaller than $b$.
To see if $b - a$ is still a positive number? If it's negative it means that actually $a > b$.

Now you are getting it (in your second sentence above).

 

[Kernul]

What does $a < b$ mean? What is a very simple way to test if $a < b$?

Anyway, I did not claim that what you are doing is wrong; I am just saying it is not the fastest way.

To deal with your original question: in your argument you started with $a \leq b$, then got $(a+c) + (-c) \leq (b+c) + (-c)$, then canceled the "$(-c)$' on both sides to end up with $a+c \leq b+c$. However, that last step is invalid: it basically assumes what you are trying to prove, because if you set $A = a+c$ and $B = b+c$ you are going from $A -c \leq B-c$ to $A \leq B$---but you have not proved that yet.

Oh right. I can't think of any way to proceed from that part right now.

Now you are getting it (in your second sentence above).

So I should simply see if $(b + c) - (a + c) > 0$?

 

[Ray Vickson]

Oh right. I can't think of any way to proceed from that part right now.So I should simply see if $(b + c) - (a + c) > 0$?

That is the way I would do it (although I would use "$\geq$" instead of "$>$").

 

[Kernul]

That is the way I would do it (although I would use "$\geq$" instead of "$>$").

Oh yes.
Doing like that I would end up with $b + c - a - c \geq 0$ and so $b - a \geq 0$. But does this really prove what I was trying to prove? Isn't this the opposite way since I started directly with $a + c \leq b + c$, which was what I had to end up while proving?

 

[Ray Vickson]

Oh yes.
Doing like that I would end up with $b + c - a - c \geq 0$ and so $b - a \geq 0$. But does this really prove what I was trying to prove? Isn't this the opposite way since I started directly with $a + c \leq b + c$, which was what I had to end up while proving?

No: you assume $a \leq b$, which means that $b-a \geq 0$. However, $(b+c) - (a+c) = b-a,$ so we have $(b+c) -(a+c) \geq 0$, which means (by definition) that $b+c \geq a+c$. No circular arguments are involved; just elementary algebra.

 

[Kernul]

No: you assume $a \leq b$, which means that $b-a \geq 0$. However, $(b+c) - (a+c) = b-a,$ so we have $(b+c) -(a+c) \geq 0$, which means (by definition) that $b+c \geq a+c$. No circular arguments are involved; just elementary algebra.

Oh, now I get it.
Thank you very much.