show that for all integers n, n^2-2n+1>=0 ?

Answer 1

OK everyone may be right, but u said "show" None of the previous answers would get full credit on a test, so allow me to show off my discrete math skills with a proof by induction:

show that for all integers n, n^2-2n+1>=0

basis step:

let n = 0
0^2 + 2(0) + 1 = 1 > 0


induction step:

let n= k
k^2 - 2k + 1 >= 0 or (k-1)^2 >= 0
since n is any integer you must take two cases, k-1 and k+1

let n = k-1: (k-1-1)^2 >= 0?
(k-2)^2 = (k-1)^2 - 2(k-1) = k^2 - 4k + 4 = (k-1)^2 - (2k - 1) + 2
since 2k - 1 <= k^2 (a rearrangement of the given), then (k-1)^2 - (2k - 1) + 2 = (k-2)^2 >= 0

let n = k+1: (k+1)^2 - 2(k+1) +1 >= 0?,
(k+1)^2 - 2(k+1) +1 = k^2 + 2k + 1 - 2k -2 + 1 = k^2
since k^2 + 1 - 2k >= 0 (given) and k^2 > 1 for any integer of k,
then k^2 > 0

Answer 2

1^2-2*1+1=0
2^2-2*2+1=1>0
(n+1)^2-2(n+1)+1-n^2+2n-1
n^2+2n+1-2n-2+1-n^2+2n-1
2n+1-2
2n-1
gor any n>=1, 2n+1>0

n^2-2n+1>=0 for n=1; n=2.

We just demonstrated the if this is true for n, it must also be true for n+1. therefore, since it is true for n=1, it must be true for n=2, 3, 4......∞

Answer 3

LHS = (n-1)^2

this being a perfect square so for all real n integer is subdomain it is >= 0

Answer 4

n²-2n+1 = (n-1)², and the square of any real number is greater than or equal to zero.

Answer 5

factorize ( n-1)(n-1) =0
equalizing each by 0
n-1=0 add one to both
n=1
so n which satisfies the equation = 1
s.s.={1}

Answer 6

where does the function f:N->R, f(n) = n^2 -2n achieve its global minimum?