Cancelling fractions

[Probability]

I have a fraction (3/4)^3 and I work out the answer to be 27/64. I initially assume that the answer is not in its lowest terms so I start to try and reduce the answer to what I think is its lowest terms.

I decide to divide 27 by 3 and 64 by 8 which gives 9/8. I then decide that I can cancel this fraction further dividing 9 by 3 and 8 by 4 = 3/2.

I find that when I carryout this calculation using the calculator that the answer is 27/64. This then leads me to understand that the fraction cannot be cancelled any further, hence I now think that cancelling fractions can only be carried out correctly if both the numerator and denominator are cancelled by the same number, i.e. 15/20 means 5 from 15 = 3 and 5 from 20 = 4, therefore 3/4 is the lowest term when cancelled.

Is this the standard when cancelling fractions and or is there variations to this method?

I think I've got it now thanks. When cancelling fractions whatever number I choose for the numerator must go into the denominator. Hence say 4/8 would mean 4 from 4 = 1 and 4 from 8 =2 therefore 1/2 is the lowest term.

 

[Steven G]

At the end you have it correct.

If you use the method at the beginning then EVERY fractions reduces to 1! In your example you got 27/64 = 3/2. But why stop there? Why not divide the 3 by 3 and the 2 by 2 and get 1/1=1
Why not from the very beginning divide 27 by 27 and 64 by 64 to get 1/1=1? Answer, because this is nonsense. Every fraction can not equal 1. We know that 7 = 7/1 =1? I think not.

It is good that you have it correct now!

 

[Dr.Peterson]

Many of us have stated repeatedly that the word "cancel" should be outlawed. This is why they say that.

I say that canceling is a useful concept, but ONLY if you understand exactly what it means in a particular context. You must have a canceling license! And you have failed the written test. (Well, okay, you sort of passed the retest.)

In a fraction, canceling means "divide the numerator and the denominator by the SAME NUMBER", or, equivalently, "remove the SAME FACTOR from the entire numerator and the entire denominator".

It absolutely does NOT mean "divide the numerator and denominator each by SOME number, whatever you feel like".

And what you need to fully understand in order to earn your canceling license is why this works:

The number 15/20 can be simplified because we can pull out a factor equal to 1:

[MATH]\frac{15}{20} = \frac{3\times 5}{4\times 5} = \frac{3}{4}\times\frac{5}{5} = \frac{3}{4}\times 1 = \frac{3}{4}[/MATH]​

What you tried to do with [MATH]\frac{27}{64}[/MATH] would not look like that; you would be pulling out a fraction that is not equal to 1, so it can't be ignored.

But you didn't have to do any of that. You started with [MATH]\left(\frac{3}{4}\right)^3[/MATH], and if the result could be simplified, then so could the original. Since obviously [MATH]\frac{3}{4}[/MATH] can't be simplified, neither can [MATH]\frac{27}{64}[/MATH].

 

[pka]

I have a fraction (3/4)^3 and I work out the answer to be 27/64. I initially assume that the answer is not in its lowest terms so I start to try and reduce the answer to what I think is its lowest terms. I decide to divide 27 by 3 and 64 by 8 which gives 9/8. I then decide that I can cancel this fraction further dividing 9 by 3 and 8 by 4 = 3/2

If \(\{j,k,n\}\subset\mathbb{Z}^+\) and \(\text{GCF}(j,k)=1\) then \(\text{GCF}(j^n,k^n)=1\)

 

[Probability]

Thank you very much for your replies. Every day is a learning curve. Some of you or all of you more than likely do this for a living, but please remember I'm not a mathematician, therefore have very little in the way of experience. I'm just an interested person who likes learning...

 

[Probability]

If \(\{j,k,n\}\subset\mathbb{Z}^+\) and \(\text{GCF}(j,k)=1\) then \(\text{GCF}(j^n,k^n)=1\)

I'd need to get the books out for that, but truthfully at this time, your work is above my pay grade. I'm just an enthusiastic learner, but thank you for your post.