stray spindle
I was struggling on question number 3 in the attached image. I found (an) answer but I don't believe it is correct because my professor stated that our proof should end with 2^(n-1)*k^n multiplied by 2. I am not sure where I am going wrong or how to get that.
wheat flumeBOT
stray spindle
Also my answer was (4k)^n and i dont think it is appropiate to say that exists in 2Z directly
I think I need more proof between that and x^n exists in 2Z
but I am not sure what
pallid meteor
x^n = (2k)^n
stray spindle
I did it again and got that x^n=2^n*k^n
pallid meteor
yeah
pallid meteor
separates a 2
stray spindle
so I would end my proof with (2x)^n exists in 2Z?
pallid meteor
$$x^n=2* (2^{n-1}*k^n)$$
undone hullBOT
Adler
pallid meteor
do you know how to get x^n=2^n*k^n?
stray spindle
So I would set the left side equal to (2k)^n? and then solve?
pallid meteor
No, we leave x^n because that's what we're trying to prove
stray spindle
What is the substitution for then? Am I plugging it into k^n and pulling a two out of both terms>
pallid meteor
Would be like:
x^n =x^n
x^n = (2k)^n sustitution
stray spindle
so is my current work wrong then?
pallid meteor
I think it is or maybe you just took too many steps.
stray spindle
because rn I am at x^n=2^n*k^n
and i dont understand how to get from there to what we are trying to get
pallid meteor
like in F2 2^a*2^b =2^(a+b)
but the other way around, instead of putting together and adding them, we are going to separate them and subtract
stray spindle
I saw that to get n-1 in the exponent for two
but I was confused on how to keep k^n
cuz if i undistrube a 2 in this equation doesnt that make the k^n not k^n
pallid meteor
We don't need to factor, everything is multiplying.
That would be if we were adding
2^2 *4 =2*(2*4) =16
not 2*(2*4/2) = 8
pallid meteor
although it doesn't really matter if we change k, the important thing is that it has the form of 2*(something)
stray spindle
isee
that makes sense
then 2(2^(n-1)*k^n) exists in 2Z because it is in the form of 2(smth)
pallid meteor
yeah
stray spindle
okay thank you!
astral heath
x=2k
//because n is a Natural number)
2^(n-1) is an integer
k^n is an integer
[2^(n-1) * k^n] is an integer
x^n = (2k)^n= 2^n * k^n = 2[2^(n-1) * k^n]
2^n * k^n is an even integer
stray spindle
I was looking at my assignment and just realized I had to format it a specific way so I ended up with this
and I am not sure how to end it
I haven't used F4 so I assume it has something to do with F4 but I don't see how that validates the statement
astral heath
and I am not sure how to end it
should be x^n = 2 * [2^(n-1) * k^n]
wheat flumeBOT
I was struggling on question number 3 in the attached image. I found (an) answer...
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stray spindle
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