Product Of Slopes Of X,y Axes=0

  • Thread starter Thread starter extreme_machinations
  • Start date Start date
  • Tags Tags
    Product
AI Thread Summary
The discussion revolves around the slopes of the x-axis and y-axis, with participants debating the mathematical properties of these lines. The slope of the y-axis (x=0) is identified as infinite, while the slope of the x-axis (y=0) is zero, leading to confusion regarding the product of their slopes. Participants clarify that the product of slopes for perpendicular lines should equal -1, and discuss the limits involved in approaching infinity. Misunderstandings about mathematical concepts, particularly regarding multiplication by zero and infinity, are addressed, emphasizing the importance of accurate interpretation. The conversation concludes with a mix of acknowledgment and playful banter about the complexities of mathematical reasoning.
extreme_machinations
Messages
54
Reaction score
0
This Is Strange !

Slope Of The Line X= 0[y- Axis] Is =+1
Slope Of The Line Y=0 [x-axis] Is= 0
Product Of The Slopes = 0

Now Product Of Two Perpendicular Lines Should Be = - 1

Is There A Contradiction ??
 
Mathematics news on Phys.org
It's funny how both the y-axis and the line y=x have the same gradient according to you :-p
 
well ,
its even funnier how lack of reading skills can hamper a persons understanding .

if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0
 
extreme_machinations said:
well ,
its even funnier how lack of reading skills can hamper a persons understanding .

if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0
I don't disagree with you, but do you know what the gradient of the line x=0 (the y-axis) is?

If you work that out you will understand where you went wrong. Please don't be so offended I was just lightly trying to point out your mistake.
 
This reminds me of the "proof" that 1 = 2. The mistake involved dividing by zero or something. ;)
 
extreme_machinations said:
well ,
its even funnier how lack of reading skills can hamper a persons understanding .

if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0

Good advice. You should take it yourself. Zurtex did not say anything about "the line y=0, which you would agree is the x-axis". He specifically referred to the "y-axis" and his point was that its slope is not 1!
 
The slope of the y-axis (x=0) is infinite. And, as we all know, infinity times zero equals negative 1. :)
 
oopsey !
tan 45 syndrome !
sorry for that ,

but i thought anything multiplied by 0 is = 0
infinity multiplied by zero being -1 is brand new information to me ,can anyone tell me how this so ??
 
extreme_machinations said:
oopsey !
tan 45 syndrome !
sorry for that ,

but i thought anything multiplied by 0 is = 0
infinity multiplied by zero being -1 is brand new information to me ,can anyone tell me how this so ??
Infinity multiplied by 0 is not -1, heck it doesn't even make sense. But what you can do is work out the limit of the 2 sides when multiplied by each other, which goes something like this:

\lim_{x \rightarrow \infty} -x \frac{1}{x} = -1

Now although it's true that:

\lim_{x \rightarrow \infty} -x = -\infty

And:

\lim_{x \rightarrow \infty} \frac{1}{x} = 0

It only makes sense to say:

\lim_{x \rightarrow \infty} f(x) g(x) = \left( \lim_{x \rightarrow \infty} f(x) \right) \left( \lim_{x \rightarrow \infty} g(x) \right)

If f(x) and g(x) have limits as x approaches infinity. And as we see above -x has no limit as x approaches infinity.
 
  • #10
Consider the following two lines:

Line A: y = -nx
Line B: y = x/n

The gradients of these two lines are:

Line A: gradient = -n
Line B: gradient = 1/n

Because any two lines which are perpendicular have gradients which multiply to give -1, we see immediately that line A is perpendicular to line B, for any given value of n.

Now, consider what happens in the limit as n goes to infinity. For line B we have:

\lim_{n \rightarrow \infty} y = \lim_{n \rightarrow \infty} x/n = 0

So, line B becomes the line y = 0.

We can write line A as:

x = -y/n

Taking the limit as n goes to infinity, we see that this line becomes the line x = 0.

Since we have taken the same limit in both cases, the lines A and B have remained perpendicular, and their gradients must still multiply to give -1. What we have, n terms of the gradients, is:

\lim_{n \rightarrow \infty} (-n)(1/n) = \lim_{n \rightarrow \infty} -1 = -1
 
  • Like
Likes Anukriti C.
  • #11
hey that's great conjecture !

thanks
 
  • #12
extreme_machinations said:
hey that's great conjecture !

thanks


What "conjecture" are you talking about? I didn't see any conjecture in this.
 
  • #13
I Was Just Using It In The General Sense ,not Strictly In The Mathematical Sense .

Please Ignore Whatever Does'nt Make Sense To You .
 
  • #14
What general sense then? I thought I knew what "conjecture" meant, even in general- and I don't see how it applies. Enlighten me.
 
  • #15
ok ,you win pal !
im not going to argue .
tell me what it was .

peace out ,
 
Back
Top