- #1

IDK10

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Let's start with the basic, x/0 = α. Where α is every number and decimal number from -∞ to +∞, by rearranging, we get x = 0α, x= 0, therefore only 0/0 = α. Now, we can integrate this with graphs.

Take the equations:

y=0/0,

and y=x/0, and the variant y=(x+a)/0

y=0/0:

If y=2 then there will be a horizontal line going through the points (x, 2), where x is any number on the x-axis. However, 0/0 = α, and therefore y=α, and as this means there will be infinite jorizontal lines, the entire graph will be full.

y=x/0:

Now we are dealing with a change in x. By rearrangin, we get x=0, therefore a graph of y=x/0 ≡ x=0. Another way of proving this, is that when x is greater than, or less than 0. It won't work, for example y=1/0, 0y=1, 0=1, but 0≠1, but if it is replaced with 0, y=0/0, we get the infinite vertical line from before, but is trapped at x=0 because of the change in x making other lines impossible.

y=(x+a)/0:

This time, by rearranging, we get x=-a. By using what we said before, it works the same way. For example, y=(x-4)/0. If x = 4, then we get y=0/0, therefore we get an infinite vertical line at x=4, but mot anyother line because if x/0=0 (where x≠0), it won't work.

The gradient of y=x/0:

y=x/0, is the same as y= x*1/0, and by differentiation we get dy/dx = 1/0, but 1/0 is impossible since, by rearranging, we get from 1/0=x to 1=0, but 1≠0, yet there is a gradient, otherwise it wouldn't be traveling upwards, the graphs and differentiation contradict each other.

The gradient of y=(x+a)/0 will be the same, as nothing would change to the line, apart from its translation across the x-axis. Proof:

y=(x+a)/0

x - dy/dx = 1

a - dy/dx = 0

dy/dx = (1+0)0 = 1/0.