5 real

$4*5*6*7*(-2)=5*6*7*8*(-1)$

Choose any 4 real numbers $a,b,c,d$ such that none of them is equal to zero or $-1$ .

Then the fifth number can be calculated as

$e = \frac{(1 + a) (1 + b) (1 + c) (1 + d)}{a b c d - (1 + a) (1 + b) (1 + c) (1 + d)}$

Of course, you need to avoid a choice which results in $abcd=(1+a)(1+b)(1+c)(1+d)$

I split it into choosing three numbers, and two numbers. Each set having this quality with products.

So: ab = (a+1)(b+1) = ab + a + b + 1

Which requires a+b = -1

By a similar argument:

x,y,z can be achieved with y = -z and y^2 = x + 1

Eg: let a,b,x,y,z be

1,-2,3,2,-2 with product 24

While increments: 2,-1,4,3,-1 also have product 24