Here are three problems involving pairs of numbers where the sum is an irrational number:
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Problem 1: Let \( a = \sqrt{2} \) and \( b = 3 \). What is the sum of \( a \) and \( b \)?
Solution: \( a + b = \sqrt{2} + 3 \) Since \( \sqrt{2} \) is irrational, \( \sqrt{2} + 3 \) is also irrational.
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Problem 2: Consider \( c = \pi \) and \( d = -\pi + 1 \). What is the sum of \( c \) and \( d \)?
Solution: \( c + d = \pi + (-\pi + 1) = 1 \) Since \( c \) (which is \( \pi \)) is irrational, the sum of \( c \) and \( d \) does not influence rationality, and the values \( d \) does not cancel out the irrationality of \( c \). Hence, in this case, \( d \) is rational, but since \( c \) is irrational, we can still find numbers where the sums yield an irrational number.
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Problem 3: Let \( e = \sqrt{3} + 1 \) and \( f = 2 - \sqrt{3} \). What is the sum of \( e \) and \( f \)?
Solution: \( e + f = (\sqrt{3} + 1) + (2 - \sqrt{3}) = 3 \) However, to ensure that the sum is irrational, let's change this up:
Let \( g = \sqrt{5} \) and \( h = -\sqrt{5} + 2 \). What is the sum of \( g \) and \( h \)?
Final Explanation: \( g + h = \sqrt{5} + (-\sqrt{5} + 2) = 2 \) Again, changing strategy works best in this context:
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A better formulated problem is: Let \( m = \sqrt{7} \) and \( n = 4 - \sqrt{7} \) (which clearly yields a sum using rational component) What is \( m + n \)?
\( m + n = \sqrt{7} + (4 - \sqrt{7}) = 4 \)
In essence, while irrational sums can't strictly rely on cancellation of values without thorough exploration of rational components, combining both rational/irrational numbers guarantees irreducible sums.
Just to clarify, ensure the values yield sums well beyond rational computation through sharpened irrational constructs.