# A recurring decimal and a parabola, how can they coincide?

• Oct 29th 2012, 05:30 AM
stuarttherock
Parabola Problem!
I am a high school student in the UK and have a mathematical problem i cannot get my head around.

It has been explained to me that 0.999... recurring equals 1.
It has also been explained that the gradient on a parabola will never be vertical.

If 0.999... can 'touch' 1 and equal 1

and the gradient of a parabola will get closer and closer to being vertical but never vertical

Is there a relationship between a decimal that gets closer and closer to 1, in that 0.999 is closer to 1 than 0.9 and that a parabola's gradient gets closer and closer to vertical?
And if 0.999.. recurring can equal 1 then why can't a parabola's gradient also be vertical?

Say you had the line y=0.999 and the y intercept is 0.999.
If you had the line y=0.999... recurring, the y intercept would be 0.999... recurring. So if you zoomed in and zoomed in on the graph, the y intercept would never be 1?

Thanks
• Oct 29th 2012, 05:32 PM
johnsomeone
Re: Parabola Problem!
".999..." is a symbol. The number that symbol represents is 1. That number *is* the number 1. They are the same. They are **EQUAL**!!
I don't think there's anything about that that's obviously comparable to a parabola's steepness (slope), which is always simply a well defined number at every point on the parabola.