To every pair of positive real numbers x and y , we assign a positive real number x ⋆ y satisfying the two properties that x ⋆ x = 1 and x ⋆ ( y ⋆ z ) = ( x ⋆ y ) × z (where × represents standard multiplication). Find k if 6 1 ⋆ 2 0 1 3 = k 1 .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Wonderful!
To generalize your steps, you showed that:
1. With
x
=
y
=
z
,
x
⋆
1
=
x
.
2. With
x
=
x
,
y
=
z
,
x
⋆
1
=
(
x
⋆
y
)
×
y
.
3. Hence
x
⋆
y
=
y
x
.
Out of everyone that I have showed this problem to, you are the only person that I know of to solve it without assuming ⋆ was the division operation. Take an upvote.
The new operation is division. x / x = 1 and x / ( y / z ) = z y x = y x × z
therefore k 1 = 2 0 1 3 6 1 = 3 3 1
k = 3 3
How can we prove that the only possibility is division?
Log in to reply
I don't think it is even true that the only possibility is division. Not sure how to show it though ; I guess we could exploit the uncountability of reals in some way , similar to what we do in Cauchy equation.
Log in to reply
See Rizky's solution above.
Log in to reply
@Calvin Lin – Oh yeah. I got that immediately after I wrote my comment :P Thanks.
Problem Loading...
Note Loading...
Set Loading...
Substitute x = y = z = 6 1 , we get 6 1 ⋆ ( 6 1 ⋆ 6 1 ) = ( 6 1 ⋆ 6 1 ) × 6 1 6 1 ⋆ 1 = 1 × 6 1 = 6 1 And then, substitute x = 6 1 , y = z = 2 0 1 3 , we get 6 1 ⋆ ( 2 0 1 3 ⋆ 2 0 1 3 ) = ( 6 1 ⋆ 2 0 1 3 ) × 2 0 1 3 6 1 ⋆ 1 = k 1 × 2 0 1 3 = k 2 0 1 3 Thus, we get k 2 0 1 3 = 6 1 k = 3 3