Find all functions f:N→Nf:\mathbb{N} \rightarrow \mathbb{N} f:N→N such that
f(2)=2f(2)=2f(2)=2
f(mn)=f(m)f(n) ∀ m,n∈Nf(mn)=f(m)f(n) ~ \forall ~m, n \in \mathbb{N} f(mn)=f(m)f(n) ∀ m,n∈N
f(m)<f(n) ∀ m<nf(m) < f(n) ~ \forall~ m < n f(m)<f(n) ∀ m<n
Note by Vilakshan Gupta 3 years, 6 months ago
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Since f(1)∈Nf(1) \in \mathbb{N}f(1)∈N and f(1)<f(2)=2f(1) < f(2) = 2f(1)<f(2)=2, we see that f(1)=1f(1) = 1f(1)=1. Certainly, then, f(j)=jf(j) = jf(j)=j for all 1≤j≤21 \le j \le 21≤j≤2.
Suppose now that f(j)=jf(j) = jf(j)=j for all 1≤j≤2n1 \le j \le 2n1≤j≤2n. Then f(2n+2) = f(2)f(n+1) = 2×(n+1) = 2n+2 f(2n+2) \; = \; f(2)f(n+1) \; = \; 2 \times(n+1) \; = \; 2n+2 f(2n+2)=f(2)f(n+1)=2×(n+1)=2n+2 and, moreover, 2n = f(2n) < f(2n+1) < f(2n+2) = 2n+2 2n \; = \; f(2n) \; < \; f(2n+1) \; < \; f(2n+2) \; = \; 2n+2 2n=f(2n)<f(2n+1)<f(2n+2)=2n+2 so that f(2n+1)=2n+1f(2n+1) = 2n+1f(2n+1)=2n+1, Thus we deduce that f(j)=jf(j) = jf(j)=j for all 1≤j≤2n+21 \le j \le 2n+21≤j≤2n+2.
By induction, then, f(j)=jf(j) = jf(j)=j for all positive integers jjj. There is only one function with this property.
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Thank you.
@Mark Hennings Sir please help me out with this problem too.
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Since f(1)∈N and f(1)<f(2)=2, we see that f(1)=1. Certainly, then, f(j)=j for all 1≤j≤2.
Suppose now that f(j)=j for all 1≤j≤2n. Then f(2n+2)=f(2)f(n+1)=2×(n+1)=2n+2 and, moreover, 2n=f(2n)<f(2n+1)<f(2n+2)=2n+2 so that f(2n+1)=2n+1, Thus we deduce that f(j)=j for all 1≤j≤2n+2.
By induction, then, f(j)=j for all positive integers j. There is only one function with this property.
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Thank you.
@Mark Hennings Sir please help me out with this problem too.