# SVM : Why can we set 1 in the hyperplane equation?

I am reading the Wikipedia article about SVM and there is something I don't understand. When they say:

These hyperplanes can be described by the equations

$$wx - b=1$$ and

$$wx - b=-1$$

I was wondering where does the +1 and -1 come from?

I found two papers which explain that this is an arbitrary choice:

We can write the following equations for the support hyperplanes:

$$w^T x = b + \delta$$ $$w^T x = b − \delta$$ We now note that we have over-parameterized the problem: if we scale w, b and $\delta$ by a constant factor $\alpha$, the equations for x are still satisfied.

To remove this ambiguity we will require that $\delta$ = 1, this sets the scale of the problem, i.e. if we measure distance in millimeters or meters

Source

but I don't understand what he means when he says "this sets the scale of the problem"

and

Note that if the equation $f(x) = wx + b$ defines a discriminant function

(so that the output is > $sgn(f(x))$),

then the hyperplane $cwx + cb$ defines the same discriminant function for any $c > 0$.

Thus we have the freedom to choose the scaling of $w$ so that $min_{x_i} |wx_i + b| = 1$.

Source

but I don't understand why he introduces $min_{x_i} |wx_i + b| = 1$.

My understanding is that we can do it because variables $w$, $b$ and $\delta$ are kind of linked together.

Can we change the Wikipedia definition and say

These hyperplanes can be described by the equations $$wx - b= 2$$ and $$wx - b=- 2$$

or is this incorrect and so we must say that :

These hyperplanes can be described by the equations $$2wx - 2b= 2$$ and $$2wx - 2b=- 2$$

Could you clarify this for me?

Very simply, if you choose any number other than 1, you can simply scale it away again. Consider $$\mathbf{w}\cdot\mathbf{x}-b=\pm\delta$$ Now, divide both sides of the equation by $\delta$, and we get $$\left(\frac1\delta\mathbf{w}\right)\cdot\mathbf{x}-\frac{b}{\delta}=\pm1$$ Which means that we can define $\mathbf{\hat w}=\frac1\delta\mathbf{w}$ and $\hat b=\frac{b}{\delta}$, and we have $$\mathbf{\hat w}\cdot\mathbf{x}-\hat b=\pm1$$ And because the goal is to minimise $\|\mathbf{w}\|$ (in order to maximise the size of the margin, $\frac2{\|\mathbf{w}\|}$), it doesn't matter if we scale $\mathbf{w}$ by some constant such as $\delta$ first - and so, we can use $\mathbf{\hat w}$ in place of $\mathbf{w}$, and the choice of $\delta$ is irrelevant (aside from needing to be a positive real number).
As for why it "sets the scale" of the problem, think of it this way: changing $\delta$ would change the scaling of $\mathbf{w}$ (that is, choosing $\delta=2$ would make $\mathbf{w}$ twice as big, for example). And so, just as changing $\delta$ changes the scale, so too does setting $\delta$ set the scale - it keeps it fixed, rather than having it vary from instance to instance.