I What does negative ratio mean in mathematics?

AI Thread Summary
Negative ratios in mathematics arise when the signs of the dividend and divisor differ, particularly in the context of external division. When dividing a line segment externally, the point dividing the segment can yield a negative ratio if it lies outside the endpoints, as the direction of the vectors changes. The discussion emphasizes that while distances themselves cannot be negative, the ratios derived from vector directions can be. The correct formulas for internal and external division are crucial for accurately representing these ratios. Understanding the vector nature of the division clarifies why negative ratios occur in this mathematical context.
parshyaa
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Whats the difference between (4/3) and -(4/3)

Why ratio in external division is negative?

I have answer for how it is negative but not why?
X = (mx2+nx1)/(m+n)

⇒ (mx2+nx1) = X(m+n)
⇒m(x2-X) = n(X-x1)
⇒(m/n) = (X-x1)/(x2-X)

From the above equation we can conclude that when division is external then X must be greater than x1 and x2, therefore ratio becomes negative
But what's the purpose of negative ratio.
 
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A ratio is negative when the divisor and dividend are of different sign.
 
hilbert2 said:
A ratio is negative when the divisor and dividend are of different sign.
Distance can never be negative then why external division gives negative ratio.
 
You can't calculate a positive definite distance without having an absolute value expression ##|x_1 - x_2 |## or a Pythagorean type formula that has squares inside a square root.
 
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What do you mean by "external" division? Divisions have a numerator and a denominator. If their signs are different, the result of the division is negative.
parshyaa said:
X = (mx2+nx1)/(m+n)
If X>x2 and X>x1, then either m or n has to be negative. So what?
 
mfb said:
What do you mean by "external" division? Divisions have a numerator and a denominator. If their signs are different, the result of the division is negative.If X>x2 and X>x1, then either m or n has to be negative. So what?
In my above example there are 3 points let us name them P(x1,y1), Q(x2,y2), and let a point R(X,Y) lies between P and Q and divides PQ in the ratio PR/QR = m/n, as given in diagram.
download.jpg

Here R divides PQ internally in the ratio (m/n) when R goes along PQ then it divides PQ externally in the ratio m/n, but when it goes along PQ its cordinate X becomes greater than x1& x2, and y becomes greater than y1 & y2, and by using my above reasoning we get to know that m/n becomes negative, how can the ratio of distance become negative
 
In your diagram, (m/n) = (X-x1)/(x2-X) is arranged so that it is positive as long as R is between P and Q. The "distances" X-x1 and x2-X are really vectors, not distances, because the direction counts. As soon as R goes outside of the [P,Q] segment, one of the directions changes and the ratio becomes negative.
 
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You are using wrong formula, the right one is ##x = {mx_2 - nx_1 \over m -n}##. The rest is explained by fact checker.
 
Buffu said:
You are using wrong formula, the right one is ##x = {mx_2 - nx_1 \over m -n}##. The rest is explained by fact checker.
Hey i know that this is the formua for external division, but i used the above formula to show that how ratio m/n becomes negative when point R goes along PQ and divides PQ externally. By using your formula we can't show that m/n is negative, check it.
 
  • #10
Sorry to all for this question, i got the idea, i just forget the modulus, (x-x1)/(x2-x) = |m/n|
Therefore [m/n] = ±[(x-x1)/(x2-x)] = , when its negative it will divide externally otherwise internally
 
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  • #11
parshyaa said:
Hey i know that this is the formua for external division, but i used the above formula to show that how ratio m/n becomes negative when point R goes along PQ and divides PQ externally. By using your formula we can't show that m/n is negative, check it.

Why do you want to show m/n is negative purposely ?
 
  • #12
Buffu said:
Why do you want to show m/n is negative purposely ?
Hey that's not purposely, m/n is always negative in external divison of line, just read the concept
 
  • #13
There are better ratios to use that are useful for interpolating between the endpoints and also for extrapolating beyond the endpoints.
 
  • #14
parshyaa said:
Hey that's not purposely, m/n is always negative in external divison of line, just read the concept

It is negative because you need to use the correct formula not the wrong one.

For a ratio (-m/n), the division is ##-mx_2 + nx_1 \over n - m## compare it with ##x_1n - mx_2\over n -m##(I applied the formula for m/n).
Both are same thing, you are just creating a fuss over nothing.
 
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  • #15
Buffu said:
For a ratio (-m/n), the division is ##-mx_2 + nx_1 \over n - m## compare it with ##x_1n - mx_2\over n -m##
How can you take ratio of two distance as (-m/n) and substitute in the formula, ratio of distance can't be negative.
 
  • #16
parshyaa said:
How can you take ratio of two distance as (-m/n) and substitute in the formula, ratio of distance can't be negative.
Why not?
 
  • #17
Svein said:
Why not?
How can you divide a line in the ratio -(4/3)
 
  • #18
parshyaa said:
How can you divide a line in the ratio -(4/3)

Yes you can.

Direction matters. When you write ##m/n## it means you are preserving the direction. To only account for distance you should write ##|m/n|## and then tell explicitly what you need to do with this ratio (divide the segment externally or internally) and then use the correct formula accordingly (See the derivation of those formulae). The benefit in writing ##m/n## is that you don't need to worry about that extra bit of information (nature of division) explicitly.

Isn't this what @FactChecker said ?
 
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  • #19
Look ...it's important to understand what section formula actually has to say...
When we say c divides AB vector then it actually means that (AC vector)/(CB vector)=k (some constant)
As a,b,c are collinear points,the k is a constant...according to collinear dependency
Now,for internal division it becomes positive as numerator and denominator (vectors) have same direction...and then by similar argument, the ration for external division will be negetive as the vectors will have opposite direction on same support
 

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