solve for n∈N (n−1)!+1=n^2


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 156951 by mr W last updated on 17/Oct/21

$s o l v e f o r n \in N$ $(n - 1)! + 1 = n^{2}$
Commented by mr W last updated on 17/Oct/21

$t h a n k s s i r!$ $i f n = 1, t h e n (n - 2)! - (n + 1) i s n o t$ $d e f i n e d . t h e r e f o r e n \neq 1 .$
Commented by Rasheed.Sindhi last updated on 17/Oct/21
$(n−1)!+1=n^2 (n−1)!−(n^2 −1)=0 (n−1){(n−2)!−(n+1)}=0 n=1 ∣ (n−2)!−(n+1)=0 But n=1 is false because it doesn′t fulfill the original.Why is it false? (n−2)!=n+1 ;n≥2 2≤n≤4: (n−2)!<n+1 n=5: (5−2)!=5+1✓ n>5: (n−2)!>n+1 n=5$
$(n - 1)! + 1 = n^{2}$ $(n - 1)! - (n^{2} - 1) = 0$ $(n - 1) {(n - 2)! - (n + 1)} = 0$ $n = 1 ∣ (n - 2)! - (n + 1) = 0$ $B u t n = 1 i s f a l s e b e c a u s e i t {d o e s n}^{'} t$ $f u l f i l l t h e o r i g i n a l . W h y i s i t f a l s e ?$ $(n - 2)! = n + 1; n ⩾ 2$ $2 ⩽ n ⩽ 4 : (n - 2)! < n + 1$ $n = 5 : (5 - 2)! = 5 + 1 ✓$ $n > 5 : (n - 2)! > n + 1$ $n = 5$
Commented by Rasheed.Sindhi last updated on 18/Oct/21

$\overset{Thanks}{mr W Sir}! {You}^{'} ve always helped me .$ $. . . I thought it only matter of the$ $definition : 0! = 1$ $If it were defined 0! = 0, the solution$ $x = 1 were not false, although the$ $other factor be undefined for it .$ $Afterall^{'} 0! = 1^{'} is a definition which$ $is made for some necessities . {What}^{'} s$ $your openion sir ? Your openion matters$ $to me .$
Commented by mr W last updated on 18/Oct/21

$i t h i n k i f 0! i s d e f i n e d, t h e n 0! m u s t$ $b e e q u a l t o 1 . w e c a n n o t f u r t h e r$ $d e f i n e 0! = 0, s i n c e o t h e r w i s e w e$ $w o u l d g e t 1! = 1 \times 0! = 1 \times 0 = 0,$ $2! = 2 \times 1! = 0 e t c .$
Commented by Rasheed.Sindhi last updated on 18/Oct/21

$Y e s S i r, {y o u}^{'} r e r i g h t!$
Commented by mr W last updated on 18/Oct/21
$0! is defined! so it is ok with (n−1)!−(n^2 −1)=0 and you can put n=1 into it. but if you factorise it to (n−1){(n−2)!−(n+1)}=0 (n−2)! must be defined, that means n−2≥0.$
$0! i s d e f i n e d! s o i t i s o k w i t h$ $(n - 1)! - (n^{2} - 1) = 0$ $a n d y o u c a n p u t n = 1 i n t o i t .$ $b u t i f y o u f a c t o r i s e i t t o$ $(n - 1) {(n - 2)! - (n + 1)} = 0$ $(n - 2)! m u s t b e d e f i n e d, t h a t m e a n s$ $n - 2 ⩾ 0 .$
Commented by Rasheed.Sindhi last updated on 18/Oct/21

$T han k s very much!$ $B u t I t h i n k n ⩾ 2 i s o n l y f o r o t h e r$ $v a l u e s o f n .$ $S i r I o n l y t a l k a b o u t s u p p o s i t i o n .$ $I f 0! = 0, t h e n . . .$ $T h e n t h e s o l u t i o n m a y c o n t a i n 1$ $o r n o t ?$
	Answered by mindispower last updated on 17/Oct/21
	$⇔n^2 −1−(n−1)!=0,n≥1,n=1not solution ∀n,N−{0,1}=E ⇔(n−1)(n+1−(n−2)!)=0 (n+1)=(n−2)! n≥5 (n−2)!≥n+1...? n=5 3!≥6 true suppose ∀n≥5 (n−2)!≥n+1,⇒(n−1)!≥n+2 (n−1)!=(n−2)!.(n−1)≥(n+1)(n−1)=n^2 −1 n.n−1≥5n−1=4n+n−1≥20−1+n=n+19>2n+2 ⇒(n−1)!>n+2 ⇒∀n≥5 (n−2)!≥(n+1),∀n>6 (n−2)!>n+1 just try n∈{2,3,4,5) 2⇒1+1=2^2 ,n=3⇒3=9,n=4⇒7=16 n=5⇒25=5^2 ⇒n=5$
	$\Leftrightarrow n^{2} - 1 - (n - 1)! = 0, n ⩾ 1, n = 1 n o t s o l u t i o n$ $\forall n, N - {0, 1} = E$ $\Leftrightarrow (n - 1) (n + 1 - (n - 2)!) = 0$ $(n + 1) = (n - 2)!$ $n ⩾ 5$ $(n - 2)! ⩾ n + 1 . . . ?$ $n = 5 3! ⩾ 6 t r u e$ $s u p p o s e \forall n ⩾ 5 (n - 2)! ⩾ n + 1, \Rightarrow (n - 1)! ⩾ n + 2$ $(n - 1)! = (n - 2)! . (n - 1) ⩾ (n + 1) (n - 1) = n^{2} - 1$ $n . n - 1 ⩾ 5 n - 1 = 4 n + n - 1 ⩾ 20 - 1 + n = n + 19 > 2 n + 2$ $\Rightarrow (n - 1)! > n + 2$ $\Rightarrow \forall n ⩾ 5 (n - 2)! ⩾ (n + 1), \forall n > 6 (n - 2)! > n + 1$ $j u s t t r y n \in {2, 3, 4, 5)$ $2 \Rightarrow 1 + 1 = 2^{2}, n = 3 \Rightarrow 3 = 9, n = 4 \Rightarrow 7 = 16$ $n = 5 \Rightarrow 25 = 5^{2}$ $\Rightarrow n = 5$
	Commented by mr W last updated on 17/Oct/21

	$t h a n k s s i r!$
	Commented by mindispower last updated on 18/Oct/21

	$w i t h e p l e a s u r s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum