Relations and Functions
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New answer posted
a month agoContributor-Level 10
[x]² + 2 [x+2] - 7 = 0
⇒ [x]² + 2 [x] + 4 - 7 = 0
⇒ [x] = 1, -3
⇒ x ∈ [1,2) U [-3, -2)
New question posted
a month agoNew answer posted
a month agoContributor-Level 9
S = {n ∈ N | [n i; 0 1] [a b; c d] = [a b; c d] ∀a, b, c, d ∈ R}
[na+ic, nb+id; c, d] = [a, b;c, d]
This must be an identity matrix. n=1. The question seems to have typos. Let's follow the solution logic.
The solution implies the matrix must be [1 0; 0 1] or [-1 0; 0 -1]. And n must be a multiple of 8.
2-digit multiples of 8 are 16, 24, ., 96. Total 11 numbers.
New answer posted
a month agoContributor-Level 10
α+β=64; αβ=256=2?
(α³/β? )¹/? + (β³/α? )¹/? = (α+β)/ (αβ)? /? = 64/32=2.
New answer posted
a month agoContributor-Level 10
f (1)=1.
f (4)=f (2)²=1 or 4.
f (6)=f (2)f (3).
Possible functions determined by values at primes: f (2), f (3), f (5), f (7).
f (2) can be 1 or 2. f (3) can be 1 or 3. f (5)=1,5. f (7)=1,7.
If f (m)=m, f (mn)=mn. One function. f (x)=1 is another.
What if f (2)=1, f (3)=3? f (6)=3.
New answer posted
a month agoContributor-Level 10
If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.
New answer posted
a month agoContributor-Level 10
∑? [ (-1)? n / 2]
= [8/2] + [-9/2] + [10/2] + [-11/2] + . + [-99/2] + [100/2]
= 4 - 5 + 5 - 6 + . + (-50) + 50
= 4
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