confusion related to elementary operation on numbers

Question

Let's take for example an fraction: $\dfrac{1+4}{2-4}$ and another fraction $\dfrac{1*4}{2*4}$. In the second fraction we can cancel 4 from both numenator and denominator but on the first we cannot do so.I have learnt how to do this but I do not have exact idea why it happens. Will anybody explain to me as clearly as possible as why that happens?

Answer 1

One way to think about it: when we cancel a common factor, we are dividing the top and the bottom by the same number. Now, division is the opposite of multiplication. Thus: $$\frac{(1*4)/4}{(2*4)/4}=\frac{1}{4}$$However, if we do this with your other example, we get:$$\frac{(1+4)/4}{(2-4)/4}$$It's incorrect to cancel, because $(1+4)/4\neq1$ and $((2-4)/4\neq2$; the division by 4 does not cancel out the addition or subtraction of 4.

We can also look at it pictorially. I'll change your minus to a plus, to make it easier to draw one of the figures.

First let me illustrate the equation $\dfrac{1*4}{2*4}=\dfrac{1}{2}$:

On the left, we have a rectangle, half of which is shaded. On the right, we have a big rectangle made up of 4 copies of the rectangle on the left. The fraction that is shaded is the same on both the left and the right.

Now let's look at the inequality $\dfrac{1+4}{2+4}\neq\dfrac{1}{2}$.

On the left we have the rectangle, half shaded. On the right we've added 4 boxes, so the shaded fraction is $\dfrac{1+4}{2+4}$. (That is, $1+4$ shaded boxes out of $2+4$ boxes total.) As you can see, the shaded fraction is quite different.

Answer 2

Because $\dfrac{a*4}{b*4} = \dfrac{a}{b}.\dfrac{4}{4} = \dfrac{a}{b}.\dfrac{1}{1} = \dfrac{a}{b}$, this is because only the operations of multiplication and division are involved, and you can swap the order around.

But, $\dfrac{a+4}{b-4} \ne \dfrac{a}{b} +/- \dfrac{4}{4}$, this is because the operations of addition and subtraction are mixed with division and you can't swap them round.

But $a + 4 - b - 4$ only involves addition and subtraction and you can swap them around (as with multiplication and division), so that $a + 4 - b - 4$ = $a + 4 - 4 - b$ = $a + 0 - b $ = $a - b$

So remember that if a calculation only involves multiplication and division (or only addition and subtraction) you can perform the individual operations in any order, but if you mix the two types then you can't.

Answer 3

Look at the meaning of equality of fractions. You have $${a\over b} = {c\over d}$$ if and only if $$ad = bc.$$ Suppose we have $a/b = c/d$ and that $z$ is a number Then $${a + z\over b+z} = {a\over b}$$ if and only if $$b(a + z) = a(b + z)$$ Multiply this out and you have $$ab + bz = ab + az$$ or if $az = bz$. Unless $z = 0$ or $a = b$, you are sunk.

You should do the same test with $${az\over bz} = {a\over b}$$ and see what happens.