Filed under: SAT Logic, SAT Math | Tags: Andrew Turner, Archimedes, fractions, mathematics, SAT Logic
The four basic operations: Addition, Subtraction, Multiplication, and Division. Every student knows what they are and how they are used, but few students can describe how the relationship between two numbers changes as we exchange a “+ or -” for a “x or ÷”. As an example, the difference between 3+5 and 3×5 is much more than simply amount.
In the case of 3+5 we are saying that 3 and 5 are two complete separate numbers that exist independently from one another. Each has its own unique identity and we combine the two to get one value.
In the case of 3×5, however, this is not the case. 3×5 is actually one number to start with, not two numbers. It is just an illusion that, when variables are thrown into the mix, creates confusion.
The Distributive Property gives us hints of this. Consider the case where we have
n(xy+z) = nxy + nz
Notice, of course, that the n is applied to the xy only once. This is because xy is one number, just as 3×5 is one number.
Even the way the obelus (÷) looks is very telling. In the case of 3÷5, it reminds us that this is really just a fraction: 3/5.
So this is the first step we come to in response to this talk of taking fractions out of school. To take away the fraction is to take away 1/4 of arithmetic.
– Andrew Turner
2 Comments so far
Leave a comment
<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <pre> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>
I thought this was an insightful comment, yet when I told my son about it, he did not appreciate it as much and responded that you can also look at variables “a” and “b”, such that X^(a*b) can be distributed to X^a*X^b, but can’t distribute X^(a+b) in the same way.
Comment by Art February 2, 2008 @ 3:00 pmHmm…well thank you very much for engaging your son in that discussion! His assumption is that X^(a*b) = X^a*X^b. Let’s see how this logic works in the real world:
Let a=2 and b=3. If we first work out X^(a*b) we get X^(2*3) which, of course is the same thing as X^(6) because 2*3 is one number: 6.
Thus, we have X^6 = X*X*X*X*X*X
Now Let us look at the case of X^2*X^3. X^2*X^3=(X*X)*(X*X*X)=X^5.
X^6 ≠ X^5
The assumption of his argument is where the issue lies and perhaps now he may better appreciate what you saw in the lesson.
In the case where we are raising a power to a power we use the proof: X^(a*b)=(X^a)^b
and
In the case where we are multiplying two terms with like bases (which means that the two terms are really one number to begin with) we use the proof: X^a*X^b=X^(a+b)
I appreciate that question very much because it is one that I have spent a long time thinking about. Keep em comin’
Andrew Turner
Comment by thinkarchimedes February 2, 2008 @ 4:44 pmhttp://www.thinkarchimedes.com