# I Check my logic

1. Jan 22, 2017

### 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?

2. Jan 22, 2017

### Comeback City

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

3. Jan 22, 2017

### micromass

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

4. Jan 23, 2017

### Staff: Mentor

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.

5. Jan 23, 2017

### Comeback City

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.

6. Jan 23, 2017

### micromass

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

7. Jan 23, 2017

### Comeback City

That being said, how would you prove it?

8. Jan 23, 2017

### micromass

Staff Emeritus
Depends on the definition and the axioms given to me.

9. Jan 27, 2017

### mikeyBoy83

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