SOLUTION: Given that {{{ sqrt( 6.4*10^n ) }}} is a rational number, what can you say about the value of n? What if both {{{ sqrt( 6.4*10^n ) }}} and {{{ root(3, 6.4*10^n ) }}} are rational?

Algebra ->  Conjunction -> SOLUTION: Given that {{{ sqrt( 6.4*10^n ) }}} is a rational number, what can you say about the value of n? What if both {{{ sqrt( 6.4*10^n ) }}} and {{{ root(3, 6.4*10^n ) }}} are rational?      Log On


   



Question 398063: Given that +sqrt%28+6.4%2A10%5En+%29+ is a rational number, what can you say about the value of n?
What if both +sqrt%28+6.4%2A10%5En+%29+ and +root%283%2C+6.4%2A10%5En+%29+ are rational? What can you say about the value of n?

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Given that +sqrt%28+6.4%2A10%5En+%29+ is a rational number, what can you say about the value of n?


sqrt%286.4%2A10%5En%29
%286.4%2A10%5En%29%5E%281%2F2%29

%28expr%2864%2F10%29%2A10%5En%29%5E%281%2F2%29

%2864%2A10%5E%28n-1%29%29%5E%281%2F2%29

%288%5E2%2A10%5E%28n-1%29%29%5E%281%2F2%29

drawing%28100%2C80%2C0%2C1%2C-1%2C1%2Clocate%280%2C0%2C8%2A10%5E%28%28n-1%29%2F2%29%29%29%29

Only integer powers of 10 are rational. So the exponent
must be an integer, say k

%28n-1%29%2F2=k

n-1 = 2k

n = 2k+1

Therefore n must be odd in order that sqrt%286.4%2A10%5En%29
be rational.

What if both +sqrt%28+6.4%2A10%5En+%29+ and +root%283%2C+6.4%2A10%5En+%29+ are rational? What can you say about the value of n?

For

root%283%2C6.4%2A10%5En%29
%286.4%2A10%5En%29%5E%281%2F3%29

%28expr%2864%2F10%29%2A10%5En%29%5E%281%2F3%29

%2864%2A10%5E%28n-1%29%29%5E%281%2F3%29

%284%5E3%2A10%5E%28n-1%29%29%5E%281%2F3%29

drawing%28100%2C80%2C0%2C1%2C-1%2C1%2Clocate%280%2C0%2C4%2A10%5E%28%28n-1%29%2F3%29%29%29%29

Only integer powers of 10 are rational. So the exponent
must be an integer, say p

%28n-1%29%2F3=p

n-1 = 3p

n = 3p+1

Therefore n must be 1 more than a multiple of 3

Now we set the exponents equal:

n = 3p+1 = 2k+1

3p = 2k

2p+p = 2k

p + p%2F2 = 2k

p%2F2 = 2k-p

The right side is an integer, so the left side must be also

Let that integer be A

Then 

p%2F2 = A 

p = 2A

Substitute in

3p = 2k

3(2A) = 2k

6A = 2k

3A = k

Now supstitute in

n = 3p+1 = 2k+1

n = 3(2A)+1 = 2(3A)+1

n = 6A+1 = 6A+1 

So n must be 1 more than a multiple of 6.

Edwin