# A relation between the law of multiplication with rate of travel and time.

I am reading the book Algebra for the practical man, chapter 1, article 6.

Uptil article-6 the author has explained the operations of addition on positive and negative numbers. Now in article-6 he is explaining(not defining) the multiplication using the concept of negative time.

• In the 4th point the author presumes that the multiplication of rate with time will give the distance moved.

In the light of this concept he concludes that $−ve×−ve=+ve$. The author states in the beginning of article-6:

We shall consider motion to the right of the starting point as positive ($+$) and motion to the left of the starting point as negative ($-$).

Then in the 4th point he says:

If the train is now at the starting point and has been traveling to the left, where was it 5 hours ago? Motion to the left $\times$ Past time $(- 40) \times (-5) = +200$

Here ($-40$) is the rate at which the train moves and ($-5$) is time elapsed.

My question is:

• As the time changes from $0$ to $-5$ the train moves from the starting point towards right in the past, shouldn't we assume the rate of train positive and conclude$(+40) \times (-5) = +200$?.
• If my interpretation is correct then does it mean that the author's assumption that
rate $\times$ time = distance is incorrect?

NOTE: Since division has not been defined uptil article 6 we cannot use the definition of division.

• It's not moving backwards in space, it's moving backwards in time. That's why the time of $5$ is inverted to $-5$ and the velocity stays the same. If you invert the velocity as well, you should expect the answer to be the same as moving with the original velocity forward in time. The acceptance of the author's assumption that Velocity x time = distance is being taken as a given to prove multiplication, so the logic must follow this rule. Which means given the premise that this rule is correct, here's how multiplication works. If you don't accept that rule, the explanation has no basis. – RandomUser Apr 25 '14 at 14:45
• @RandomUser If it moves backward in time then it will also move backward in space, won't it? Like when a video is played backward the positions of things are changed to their past positions. – user103816 Apr 25 '14 at 15:16
• Exactly. The position of them changes, and they appear to be moving backwards, but that's because time is. If you took a video of a car moving forward and rewound it, you wouldn't see the driver throw the gear into reverse. Another way to think of it is to draw a graph of position over time. When looking back in time, you're just moving left on the graph. The change of position over time (the slope of the line, which is the velocity) doesn't change when you do this. – RandomUser Apr 25 '14 at 15:34
• @RandomUser Here is the point. The velocity is defined for the positive time,that is $v=\dfrac{x_2-x_1}{t_2-t_1}$, $t_2-t_1$ is the magnitude of future time. The slope is defined from this. The problem at hand is that uptil art-6 author hasn't defined division. The author simply states:"We shall consider motion to the right of the starting point as positive (+) and motion to the left of the starting point as negative (-).". Now if we move from present to past the velocity then should be +ve acc. to author's statement. – user103816 Apr 25 '14 at 16:13
• All the down-votes this question make no sense. My suspicion is that people who don't know how to answer a question but think they should know, express themselves in that way. – Michael Hardy May 19 '14 at 20:02

The rate is negative because the train is moving toward the left.

The time is negative because it's five hours before the time we're calling zero.

Negative rate times negative time equals positive distance, and positive means it's to the right of the location we're calling zero.

You say as time changes from $0$ to $-5$, the train moves rightward. Rightward does indeed correspond to positive distance. The change in time is $-5$ and the change in distance is $+200$ therefore the rate is $$\text{rate} = \frac{\text{change in distance}}{\text{change in time}} = \frac{+200}{-5} = -40.$$ The rate is negative.

• Thankyou for the answer, but I guess you have missed the last line of my question. – user103816 May 22 '14 at 7:43

I don't understand the downvotes to this question either.

The author defines $\mbox{distance}= \mbox{rate} \times \mbox{time}$. Somewhere in his definition of rate, he states that rate is positive if moving right, negative if moving left. Perhaps what wasn't clear is that this definition also assumes that the arrow of time has been fixed: that time is increasing. Therefore, rate is actually independent of (the sign of) time. In other words, when speaking of a rate, the convention should be "with respect to increasing time." You probably know from other places that velocity is distance/time, where time is increasing (positive). The reason for all this is to establish a convention so that when talking about a car moving at 50mph, we both assume that this is defined with respect to increasing time (otherwise we always have to establish a convention, which is very annoying). Notice that the author does not need the notion of division in any of this. Again, to be precise in his defition, the author needs to mention that the rate (velocity) is with-respect-to increasing time.

Given that definition, the problem should be clear. If we reverse time, then the distance covered correctly reverses as well, so if $d=vt$, then $-d=v(-t)$.

To clarify even further, some time ago mathematicians establish that $5$ is "positive five" and $-5$ is "negative five." This is a set convention, so it's unlikely that you would think $5$ is negative. In the same way, the concept of rate is fixed with respect to increasing time. The author extends the definition of distance covered to negative time by multiplying by negative time instead, in the same way that you multiply $5$ by $(-1)$ to get $-5$.