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I'm very new to discrete math so any help or guidance would be great.

Proof: $[ a ≤ c ∧ a+b ≤ c+d] $

I know to prove partial order you have to prove reflextivity, anti symmetry, and transitivity and this is what I have so far...

Reflextivity: $R$ is reflexive if $(a,b)R(a,b)$ for all $a,b \in \Bbb Z$

Antisymmetry: ...

Transitivity: Suppose $(a,b)R(c,d)$ and $(c,d)R(a,b)$

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  • $\begingroup$ So you are proving that [a≤c∧a+b≤c+d]? What is R? How can R have anything to do with what you are trying to prove? Without knowing anything about a,b,c,d, it is impossible to prove [a≤c∧a+b≤c+d]. Would you state your question so it is coherent, so it does not require decyphering? $\endgroup$ – William Elliot Apr 23 '18 at 22:18
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HINT

So you have a relation $R$ defined on $\mathbb{R}^2$, where $$ R((a,b),(c,d)) = \{a \le c \mathrm{\ and\ } a+b \le c+d\}. $$

So let's check reflexivity.

$R((a,b), (a,b))$ requires us to check if $a \le a$ and $a+b \le a+b$. Are these statements true? So is $R$ reflexive?

Similarly check transitivity and anti-symmetry. If all 3 are true, $R$ is a partial order.

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