use strong induction to prove :?

Question

route of 2 in an irrational number. being by letting p(n) be the statment thet rout of 2 not equal n/b for any positive integer b.

Answer 1

It's interesting that you're trying to prove this using math induction. You have it stated correctly.
For n = 1, sqrt 2 not equal to 1/b since b is a positive integer and sqrt2 is obviously > 1.
Assume for n = 1, 2, . . . , k, sqrt 2 not = n/b
If sqrt2 = (k + 1)/b then
The only way I can continue is to apply the standard proof that the sqrt2 is irrational on the last statement which seems a waste of time.
If you come up with the math induction proof, I'd like to see it.
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Answer 2

i think of you're referring considered one of those evidence in which you do right here: (a million) make certain the fact is right for the 1st case. (2) assume it relatively is real for the nth case, and use this to tutor it relatively is real for the (n+a million) case. occasion: to tutor the sum of the 1st n integers is n(n+a million)/2: (a million) it works interior the 1st case, i.e. n=a million: a million(a million+a million)/2 = a million. (2) Assuming it works for the nth case, then a million+2+...+n=n(n+a million)/2. Then, for the (n+a million) case, a million+2+....+n+(n+a million) = n(n+a million)/2 + (n+a million) = (n^2 +3n +2)/2 = (n+a million)(n+2)/2. That proves the formula works.

Answer 3

That's not induction, that's assuming what you are trying to prove.