0
No
A rational number is a one that can be written as a fraction i.e a/b. where a and be are integers (whole numbers)
Consider
a/b and c/d. Where a b c and d are integers and as such rational numbers
a/b + c/d = (ad + bd)/cd
ad, bd and cd will all be integers and as such a/b + c/d will always be rational
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Can you add irrational number and a rational to get a rational number?
If an irrational number is added to, (or multiplied by) a rational number, the result will always be an irrational number.
Explain why the sum of a rational number and an irrational number is an irrational number?
Let R1 = rational number Let X = irrational number Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)
Why an irrational number plus an irrational number equal a rational?
Well, darling, when you add two irrational numbers together, they can sometimes magically cancel each other out in such a way that the sum ends up being a rational number. It's like mixing oil and water and somehow getting a delicious vinaigrette. Math can be a wild ride, honey.
Can you add three irrational numbers to get a rational number?
Yes. sqrt(2), 2*sqrt(2) and -3*sqrt(2).
If you add a rational and irrational number what is the sum?
an irrational number PROOF : Let x be any rational number and y be any irrational number. let us assume that their sum is rational which is ( z ) x + y = z if x is a rational number then ( -x ) will also be a rational number. Therefore, x + y + (-x) = a rational number this implies that y is also rational BUT HERE IS THE CONTRADICTION as we assumed y an irrational number. Hence, our assumption is wrong. This states that x + y is not rational. HENCE PROVEDit will always be irrational.
