a = [1,2,3] b = a a[0] = "Bob" c = [1,2,3] c[0] = b[0] What is the first element in c?

a = ["Bob", 2, 3] b = ["Bob", 2, 3] c = ["Bob", 2, 3] Both a and b point to the same object. Even though the first element of a was changed, b was also changed, thats why when c[0] is set to b[0], c[0] is also "Bob"

What is the linearization of $$arctan(xyz)$$ at the point $$(1,1,0)$$

The linearization of any function in n-dimension is characterized by the initial value at the particular point plus all the instantaneous change in each of the dimensions times the change. In three dimensions it's: $$L(x, y, z) = f(x_0, y_0, z_0) + f_x(x_0, y_0, z_0)(x − x_0) + f_y(x_0, y_0, z_0)(y − y_0) + f_z(x_0, y_0, z_0)(z − z_0).$$ Since $$arctan(0)$$, $$f_x(1,1,0)$$ and $$f_y(1,1,0)$$ = 0 the first three terms all equal 0. $$L(x, y, z) = f_z(x_0, y_0, z_0)(z − z_0)$$ which is just $$= \frac{x_0y_0}{1+z_0^2}(z -z_0)$$ plugging in $$(x_0,y_0, z_0) = (1,1,0)$$ $$= \frac{1}{1} (z-0)$$ $$L(x, y, z) = z$$

Assume 2 random variables $$a$$ and $$b$$ $$a = b$$ $$a^2 = ab$$ Multiply by $$a$$ $$a^2 - b^2 = ab - b^2$$ Subtract $$b$$ from both sides $$(a + b) (a-b) = b(a-b)$$ Foil $$a + b = b$$ Cancel out the $$(a-b)$$ since it appear on both sides $$2b = b$$ Since $$a=b$$ the assumption at the beginning $$\therefore2 = 1$$ Obviously, this is incorrect, but what algebraic mistake was made?

During the division step $$a-b$$. Since we assumed $$a$$ and $$b$$ were equal $$a-b$$ is equal to zero, and division by zero is undefined in mathematics. This fallacy could prove anything even 0 = 1 or 100000 = 1.