# A system of linear inequality is equivalent to a system of strict linear inequality

$a,x_i \in \mathbb{R}^p, b \in \mathbb{R}$. Consider the following two systems of linear inequality:

$$\label{eq:eq1} \begin{cases} \langle a, x_i \rangle + b > 0&\text{if }y_i = 1 \\ \langle a, x_i \rangle + b < 0&\text{if }y_i = -1 \end{cases}$$ $$\label{eq:eq2} \begin{cases} \langle a, x_i \rangle + b \ge 1&\text{if }y_i = 1\\ \langle a, x_i \rangle + b \le -1&\text{if }y_i = -1 \end{cases}$$ The question is if two systems of linear inequality are equivalent.
It is easy to see that any a and b satisfying the system in equation (2) will also satisfy the system in (1). However, how do I prove that a and b satisfying the system in (1) will also satisfy the system in (2).

Note: The statement is from here. It is Remark 1 on page 2. I just don't quite understand their explanation.

Firstly, please notice that the link you shared is not saying a point $x_i$ satisfying the first set satisfies the second. Instead, it says that if you write such a classification in the $1^{st}$ set, and if there is a feasible solution then the $2^{nd}$ set has a feasible solution too.

The reason we do such a transformation is that a linear model can't solve strictly greater/larger condition.

I am attaching a motivating real life example as an image below.

• can you please give me the source for the image? May 2, 2018 at 15:16
• It's my work in progress. I just wanted to show you a practical application. You can always use this trick for linear models with similar settings. May 2, 2018 at 21:08

They are not equivalent. There is no assumption about $b$. So $⟨a,x_i⟩+b$ can be i.e. $0.5$ when $y_i = 1$ and not fulfill the second condition. I think you should give more details about $a$ and $b$ to have an exact answer.

• I updated my question. May 2, 2018 at 1:32
• Adding a new answer. Check that. May 2, 2018 at 10:01