Verified answer

(1-n)/(1-n²)<0.01

(1-n)/(1-n)(1+n)<0.01

1/(1+n)<0.01

the common denominator is (1+n)²

as it is an inequality

1+n<0.01(n+1)²

1+n<0.01(n²+2n+1)

0.01n²-0.98n -0.99>0

(n+1)(0.01n-0.99)>0

this gives us

n<-1 or n>99

1-n/(1-n^2) < 0.01 {multiply by 100 to elimimate the decimal}

100(1-n) / (1-n^2) < 1 {multiply by 1-n^2 to elimilate the fraction}

**Note that 1-n^2 = (1-n)(1+n) and is in the denominator so n can't be 1 or -1.

100-100n < 1-n^2 {move everything to the left to create a quadratic}

n^2 - 100n + 99 < 0

(n-99)(n-1) < 0

The quadratic 2 steps before has a coefficient on the squared term that is positive, so this is a parabola opening up with zeros at n = 1 and 99.

Then between 1 and 99, this parabola would be less than zero.

So n < 1 OR n > 99

The other criteria we are given is that n ∈ N.

So this depends on how you define N.

If N = {0,1,2,3...}, then the answer is {0} + {100, 101, 102...}

If N = {1,2,3...} (not including zero), then the answer is {100, 101, 102...}

So use your appropriate definition of N for your final solution.

I looked online at wikepedia and Wolfram and There was no consensus as to which of the above was accurate.

I hope this helps

(1 - n)/(1 - n^2) < 0.01

(1 - n)/((1 - n)(1 + n)) < 1/100

1/(1 + n) < 1/100

1 + n > 100

n > 99

If this is (1-n) / (1-n^2) < 0.001, then we have: 1/(1+n) < 1/1000, so n>1000