Let, R be a relation defined on N×N such that (a,b)R(c,d)⟺ad(b+c)=bc(a+d);a,b,c,d∈N. Is the relation R Transitive?

To determine if the relation R defined on N×N (the set of natural numbers) is transitive, we need to check whether for any three ordered pairs (a, b), (b, c), and (a, c) that satisfy the relation, (a, b)R(c, d) and (c, d)R(e, f), it implies that (a, b)R(e, f).

In other words, we want to see if the following is true for any natural numbers a, b, c, and d:

  1. If ad(b+c) = bc(a+d) and cd(e+f) = de(c+f), does it imply that ad(b+c) = de(c+f)?

Let’s analyze this condition:

Starting with ad(b+c) = bc(a+d) and cd(e+f) = de(c+f), we can isolate the terms containing a, b, c, d, and e:

  1. ad(b+c) = bc(a+d) (Equation 1)
  2. cd(e+f) = de(c+f) (Equation 2)

Now, let’s focus on Equation 1:

ad(b+c) = bc(a+d)

We can expand both sides:

abd + adc = abc + bcd

Now, let’s subtract bc(a+d) from both sides:

abd – abc = bcd – adc

Factor out common terms:

ab(d-c) = cd(b-a)

Now, let’s divide both sides by (d-c) (assuming d ≠ c) and (b-a) (assuming b ≠ a) because we are dealing with natural numbers:

ab = cd

So, from Equation 1, we can see that ab must be equal to cd for the relation R to hold.

Now, let’s return to Equation 2:

cd(e+f) = de(c+f)

Expand both sides:

cde + cdf = cde + def

Subtract cde from both sides:

cdf = def

Now, divide both sides by f (assuming f ≠ 0):

cd = de

So, from Equation 2, we can see that cd must be equal to de for the relation R to hold.

In conclusion, the relation R is indeed transitive because it satisfies the transitive property for any natural numbers a, b, c, d, and e.

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