Don't understand classification equation for hard margin SVM I am trying to get a grasp of hard margin SVMs. In the lecture I am watching the professor talks about a classification equation which when a positive sample is input, returns a value of $1$ or more; and when a negative sample is input, returns a value of $-1$ or less. The graph below shows the vector $\overrightarrow{w}$ which is perpendicular to the separating hyperplane and an arbitrary point with unknown class $\overrightarrow{u}$.



In the lecture it says that:

$$\overrightarrow{w} \cdot \overrightarrow{u_+} + b \geq 1 \textrm{ , for positive class samples}$$

$$\overrightarrow{w} \cdot \overrightarrow{u_-} + b \leq -1 \textrm{ , for negative class samples} $$

To me it seems that we should take the projection of $\overrightarrow{u}$ on $\frac{\overrightarrow{w}}{|\overrightarrow{w}|}$, since this would give the component of $\overrightarrow{u}$ in the direction of $\overrightarrow{w}$. If this component is greater than the distance to the decision boundary/hyperplane, $b$, then $\overrightarrow{u}$ is a positive sample, if it is less then it is a negative sample. In math terms

$$\frac{\overrightarrow{w}}{|\overrightarrow{w}|} \cdot \overrightarrow{u_+} - b &gt; 0$$

$$\frac{\overrightarrow{w}}{|\overrightarrow{w}|} \cdot \overrightarrow{u_-} - b &lt; 0$$

If $\overrightarrow{u}$ lies on the decision boundary then the above expressions will be equal to 0.

If $\overrightarrow{u}$ lies on a support vector hyperplane, then the above expressions will be equal to $m$ or $-m$ for positive and negative samples respectively, where $m$ is the margin from the decision boundary to any support vector.

I don't understand the lecturer's equations. Why is the value for positive sample classification $\geq 1$? Why is the value for negative sample classification $\leq -1$? Furthermore, what does $b$ represent in the lecturer's equations? 

Basically, I have trouble understanding the proposed equations, but they should be doing the same thing as mine.
Very good question. Its worth understanding following concepts about SVM to understand about your point  


SVM get trained from the training dataset in such a way that for every instance Xi in dataset
W.Xi + b &lt;= -1 if Xi belongs to -ve class
W.Xi + b >=1 if Xi belongs to +ve class
This also means that for edge instances of +ve and -ve classes the value will be 1 and -1 respectively. This is what it means by making the street as wide as possible.
Above condition is not necessarily true with new unseen instance 'U' because it hasn't been seen by SVM algorithm yet so it can also be inside the street separating the classes
Above equation also implies that the edge instances are separated by 2 units (whatever the unit is). To make the street gap 2 unit, W not give projection of Xi perpendicular to the street but it also scales it to make the gap 2 units. This means W is not necessarily a unit vector.
In other word we are not comparing the component of U in direction perpendicular to the street but scaled version of it so that the street is always 2 units wide


This is intuition based explanation of SVM but mathematically also above equation is the constrain of the optimization problem needed to solved for SVM