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Sagot :
Answer: Always true
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Explanation:
We can prove this by contradiction.
Let's say
- A = some rational number
- B = some irrational number
- C = some other rational number
and
A+B = C
We'll show that a contradiction happens based on this.
If A is rational, then A = p/q where p,q are two integers. The q cannot be zero.
If C is rational, then C = r/s for some other integers. We can't have s be zero.
Note the following
A+B = C
B = C - A
B = r/s - p/q
B = qr/qs - ps/qs
B = (qr - ps)/qs
B = (some integer)/(some other integer)
This shows B is rational. But this is where the contradiction happens: We stated earlier that B was irrational. A number cannot be both rational and irrational at the same time. The very definition "irrational" literally means "not rational".
In short, I've shown that if A+B = C such that A,C are rational, then B must be rational as well.
The template is
rational + rational = rational
Therefore, we've shown that if A is rational and B is irrational, then C cannot possibly be rational. C is irrational.
Another template is
rational + irrational = irrational
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