Inequality proof: If a>b implies a>c then b>c

[dimitri151]

Summary:: To prove a conditional statement on a pair of inequalitites.

Mentor note: Moved from technical forum section, so the post is missing the usual fields.
I feel it should be possible to prove this but I keep getting lost in the symbolic manipulation.
Theorem: If a>b implies a>c then b>c.
Intuitively if every time a number a is greater than a number b it is also greater than a number c then b>c.
Is this correct?

 

[Mark44]

Summary:: To prove a conditional statement on a pair of inequalitites.

Mentor note: Moved from technical forum section, so the post is missing the usual fields.
I feel it should be possible to prove this but I keep getting lost in the symbolic manipulation.
Theorem: If a>b implies a>c then b>c.
Intuitively if every time a number a is greater than a number b it is also greater than a number c then b>c.
Is this correct?

The "theorem" is not true in general. Suppose a = 3, b = 1, and c = 2.
We have a > b being true, and we have a > c also true, but b < c.

 

[FactChecker]

No. It is not correct. Even if you interpret it as ($\forall a \in R, a \gt b \implies a \gt c$) implies ($b \gt c$), the statement is false when $b = c$.

 

[haruspex]

The "theorem" is not true in general. Suppose a = 3, b = 1, and c = 2.
We have a > b being true, and we have a > c also true, but b < c.

I think you have overlooked the significance of the "implies". The concept is that if b and c are such that, for all a, a>b implies a>c then it says something about the relationship between c and b.
As @FactChecker notes, the correct implication is b≥C.

 

[PeroK]

No. It is not correct. Even if you interpret it as ($\forall a \in R, a \gt b \implies a \gt c$) implies ($b \gt c$), the statement is false when $b = c$.

So, the original theorem needs to be: $$(\forall a \in \mathbb R, a \gt b \implies a \gt c) \implies (b \ge c)$$

 

[Nugatory]

Stuff like this is the reason we need and use the formal symbolic notation.

 

[dimitri151]

Ah, it's a tautology in the new form by PeroK:
$(a>b\Rightarrow a>c)\Rightarrow(b\geq c)$
$=[\neg (a>b)\vee (a>c)\Rightarrow(b\geq c)$
$=(a\leq b\vee a>c)\Rightarrow b\geq c$
$=\neg(a\leq b\vee a>c)\vee b\geq c$
$(a>b\wedge a\leq c)\vee (b\geq c)$
$=b<c \vee b\geq c$

Thanks for your help guys.

 

[dimitri151]

Goodness gracious, now I think it IS correct again! Aaarg..

Stop me where I go wrong:
$(p\rightarrow q)\rightarrow r$
$=(\neg p\vee q)\rightarrow r$
$=\neg(\neg p\vee q)\vee r$
$=(p\wedge \neg q)\vee r$
If
$p=a\geq b$
$q=a\geq c$
$r=b\geq c$
Then $(a\geq b\rightarrow a\geq c)\rightarrow b\geq c$
$=(a\geq b\wedge a<c)\vee b\geq c$
$b<c\vee b\geq c$ is a tautology, that is, is always true.

 

[haruspex]

it IS correct

What exactly is "it" here?
You started off with
$$(\forall a \in \mathbb R, a \gt b \implies a \gt c) \implies (b \gt c)$$
which is not true.
In post #7 you showed $$(\forall a \in \mathbb R, a \gt b \implies a \gt c) \implies (b \ge c)$$
And in post #8
$$(\forall a \in \mathbb R, a \ge b \implies a \ge c) \implies (b \ge c)$$

What is the strongest true version?

 

[PeroK]

It may also be instructive to look at the contraposition: $$(c < b) \implies (\exists a, c < a < b)$$ Note that you can replace either or both of the $<$ with $\le$ in the second expression.

Note also that if $c \le b$, then you can't necessarily find a similar $a$, because of the case $b = c$.