Comments for SAT Math and Logic http://thinkarchimedes.wordpress.com Think Archimedes SAT Math and Logic Sat, 02 Feb 2008 21:44:34 +0000 http://wordpress.com/ hourly 1 Comment on Arithmetic’s Optical Illusion by thinkarchimedes http://thinkarchimedes.wordpress.com/2008/02/01/arithmetics-optical-illusion/#comment-4 thinkarchimedes Sat, 02 Feb 2008 21:44:34 +0000 http://thinkarchimedes.wordpress.com/?p=7#comment-4 Hmm...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 www.thinkarchimedes.com Hmm…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
http://www.thinkarchimedes.com

]]> Comment on Arithmetic’s Optical Illusion by Art http://thinkarchimedes.wordpress.com/2008/02/01/arithmetics-optical-illusion/#comment-3 Art Sat, 02 Feb 2008 20:00:55 +0000 http://thinkarchimedes.wordpress.com/?p=7#comment-3 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. 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.

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