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vekshin1
3 years ago
14

Prove that the product of a nonzero rational number together with an irrational number is an irrational number.

Mathematics
1 answer:
melamori03 [73]3 years ago
7 0
Each time they assume the sum<span> is </span>rational<span>; however, upon rearranging the terms of their equation, they get a contradiction (that an </span>irrational number<span> is equal to a </span>rational number<span>). Since the assumption that the </span>sum of a rational<span> and </span>irrational number<span> is </span>rational<span>leads to a contradiction, the </span>sum<span> must be </span>irrational<span>.</span>
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Step-by-step explanation:

13r=117

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Step-by-step explanation:

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For what value of c is the function defined below continuous on (-\infty,\infty)?
kozerog [31]
f(x)= \left \{ {{x^2-c^2,x \ \textless \  4} \atop {cx+20},x \geq 4} \right&#10;

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if 
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</span><span>In notation we write respectively
</span>\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
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Thus these two limits, the one from above and below are equal if and only if
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That is to say, if c = -2, f(x) is continuous at x = 4. 

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-\infty, +\infty)

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Which expression has a value of −10?
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D

Step-by-step explanation:

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