Q: Determine whether the relation is reflexive, symmetric and transitive:
Relation R is the set N of natural numbers defined as,
ㅤ R = {(x, y) : y = x + 5 and x < 4}
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Answers
Answer:
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R.
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈/
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈/R. ∴R is not symmetric.
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈/R. ∴R is not symmetric.Now, since there is no pair in R such that (x,y) and (y,z)∈R, then (x,z) cannot belong to R.
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈/R. ∴R is not symmetric.Now, since there is no pair in R such that (x,y) and (y,z)∈R, then (x,z) cannot belong to R.∴R is transitive.
R={(x,y):y=x+5andx<4}={(1,6),(2,7),(3,8)}It is seen that (1,1)∈/R⇒R is not reflexive.Also (1,6)∈R. But, (1,6)∈/R. ∴R is not symmetric.Now, since there is no pair in R such that (x,y) and (y,z)∈R, then (x,z) cannot belong to R.∴R is transitive.Hence, R is neither reflexive, nor symmetric, but transitive.
Step-by-step explanation:
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According to algebra, every mathematical operation is associated with some major identities. Additive identity and multiplicative identity are the two basic algebraic identities. For rational numbers, natural numbers, whole numbers, and integers zero is the additive identity and 1 is the multiplicative identity.
