Can a be proven to be greater than or equal to c from given inequalities?

[mikeyBoy83]

Given $a,b\inℝ$ with $a≥b$ and $b≥c$, I wish to show that $a≥c$. For that we have $a-b≥0$ and $b-c≥0$. Therefore, $a-b+b-c≥0$ and so $a-c≥0$. Hence, $a≥c$. Q.E.D.

Correct me if I am wrong, but this works unless I should use different cases because of equality and inequality. What do you think?

 

[Comeback City]

Given $a,b\inℝ$ with $a≥b$ and $b≥c$, I wish to show that $a≥c$. For that we have $a-b≥0$ and $b-c≥0$. Therefore, $a-b+b-c≥0$ and so $a-c≥0$. Hence, $a≥c$. Q.E.D.

Correct me if I am wrong, but this works unless I should use different cases because of equality and inequality. What do you think?

Looks good to me. Even simpler, this can be written as a≥b≥c, and automatically, a≥c.

 

[micromass]

Even simpler, this can be written as a≥b≥c, and automatically, a≥c.

How does this prove anything. You just restate the result which needs to be proven.

 

[mfb]

How do you get from $a \geq b$ to $a-b \geq 0$ and from ($a-b \geq 0$ and $b-c \geq 0$) to $a-b +b-c \geq 0$?

At that level, you have to do everything with the definitions and axioms.

 

[Comeback City]

How does this prove anything. You just restate the result which needs to be proven.

It shows that if a is greater than/equal to b, then it must be greater than/equal to whatever b is greater than/equal to: in this case, c.

 

[micromass]

It shows that if a is greater than/equal to b, then it must be greater than/equal to whatever b is greater than/equal to: in this case, c.

You're assuming what requires to be shown: transitivity.

 

[Comeback City]

You're assuming what requires to be shown: transitivity.

That being said, how would you prove it?

 

[micromass]

That being said, how would you prove it?

Depends on the definition and the axioms given to me.

 

[mikeyBoy83]

My proof as written is incomplete. In the final proof I'd be sure to justify my assertions using definitions and axioms.