Really Real Numbers!

Algebra Level 4

$a$ and $b$ are two real numbers. Read the following statements about them.

$[1].$ If $a>b$ and none of them are equal to zero, then $\dfrac{1}{a}<\dfrac{1}{b}$ .

$[2]$ . The equation $\sin^{-1} (a)+\cos^{-1} (a)=\dfrac{\pi}{2}$ holds for all real numbers $a$ .

$[3]$ . $\dfrac{a}{b}$ is always a real number.

Which of these statements are true?

Details and assumptions :

The statements are independent. That means if according to statement $[1]$ , $a=2$ ; it applies to statement $[1]$ only.

This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

None of them are correct.

[2]

and

[3]

[1]

and

[2]

Only

[1]

2 solutions

Mursalin Habib
May 12, 2014

Consider the statements one by one.

$[1]$ is true if and only if $ab>0$ . In other words $[1]$ is true if $a$ and $b$ have the same sign. As a counter-example, let $(a, b)$ equal $(7, -6)$ . What happens?

Of course $[2]$ is not true for all real numbers $a$ . In fact if $|a|>1$ , the equation is meaningless.

$[3]$ isn't true because $\dfrac{a}{b}$ is not a real number if $b=0$ .

So none of these statements are correct.

Good one dude. Reminds me of Harry Potter 4 : Constant vigilance!

Chandrachur Banerjee - 7 years, 1 month ago

Nice explanation missed the last statement I did-not consider b=0.

Mardokay Mosazghi - 7 years, 1 month ago

Nice question. I forgot about the domain of arcsin :P

Michael Tang - 7 years, 1 month ago

Very slick in putting none of them are zero in the first and omitting it in the third... :)

kalyan pakala - 7 years ago

I misread and thought that first statement from [1] applied to [3]. Derp.

Justin Wong - 7 years, 1 month ago

Sorry. I can see how that could be a problem. I'll edit the question.

Mursalin Habib - 7 years, 1 month ago

Great One - Loved to Solve it. A great question indeed.

gulshan kumar k - 7 years, 1 month ago

Thanks! :)

Mursalin Habib - 7 years, 1 month ago

Yes. I took only (1) is correct.

Kevin Patel - 7 years ago

nice ...I like it..

Pathik Patel - 7 years ago

Good one, looked pretty simple, but did not consider the last point

Gajendra Singh - 7 years ago

only[1] is the right answer if both have the same sign , which created a misconception for me and i chose the wrong answer

Apurv Rajput - 7 years, 1 month ago

Akhilesh Vibhute
Dec 3, 2015

Clarity of concepts and purity in brain only can help one to solve such problems..... As simple as that

Really Real Numbers!

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