Are you sure inequality
True or False?
If x is a real number satisfying x 1 < 1 0 , then it must be true that x > 1 0 1 .
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10 solutions
But for x=-1 , x is not greater than 1/10 which is told that must be satisfy
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Yah, so the answer is f a l s e .
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Say if x=2 then equations hold good for both the statements thus true. Wdu say
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@Chandramouli Guha – The answer is f a l s e if there exist one case that x not satisfy it.
@Chandramouli Guha – You should learn some simple logic. The question say: x is real number, = > x satisfy statement above. As, I prove that x doesn't satisfy the statement above, the answer is f a l s e .
The only equation that needs to be satisfied is 1/x < 10, when x is a real number, and x > 1/10 is what we are trying to prove or disprove, not the other way around. X = -1 satisfies 1/x < 10, since that simplifies to -1 < 10, which is both true for the real number and the equation. However, x = -1 does not satisfy x > 1/10, since a negative number can't be greater than a positive number, therefore, the statement is false.
But it is given that x is a real number . so how can we take x = -1
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x=-1 is also a real number. Real number include negative.
Relevant wiki: One-step Linear Inequalities
Generally speaking, for non-zero reals a and b ,
a < b ⟹ a 1 > b 1 if and only if a and b are either both positive or both negative.
a < b ⟹ a 1 < b 1 if and only if a and b differ in sign.
How do you prove this ?
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Use the definition of < .
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And the Field Axioms of R , I suppose.
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@Agnishom Chattopadhyay – Yes, we need both Algebraic and Order Axioms.
It is not true for all negative x . Let say, x = − n [where n is a positive real number]
− n 1 < 10
Since, − n < 1 0 1
Therefore the given statement is FALSE.
But if x=2 then the equations hold true for both cases
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In order for a statement to be true, it must be true for all cases. Since the problem states "real numbers", then the equation must hold true for all real numbers, which it fails, since it fails for negative numbers.
The question does not ask for the range of values for x for which the inequality holds. It simply asks if the two inequalities are equivalent. This is a simple problem of balancing equations. "What you do on one side must be done on the other" - From 1/x >10, multiply both sides by x to get x/x >10x, i.e. 1 > 10x - Then divide both sides by 10 to get 1/10 > 10x/10, i.e. 1/10 > x ....which was what was asked.
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...which of course makes the correct answer True.
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Your manipulation of the inequalities doesn't work if x is negative; negative x refutes the statement.
Specifically, multiplying both sides by x would create either x/x>10x for positive x, or x/x<10x for negative x.
Since the statement isn't true for negative x, the answer must be false.
It's true only if x is positive, but not if x is negative
Yep. So, in general, it is not true.
Whenever you multiply or divide both sides by a negative number , the inequality flips direction
Yes, this is a very common mistake that all of us often commit when simplifying inequalities.
If in any case x is not a positive this theory will be proven false !
The statement \frac {1}{-1} <10 is true.
The statement -1 > \frac {1}{10} is false.
\frac{1—10x}{x}<0 is satisfied only if x>\frac{1}{10}.
1/-2 = -0.5. This is not bigger than 1/10.
"if x 1 < 10 then it is must be true that x > 1 0 1 "
- This statement is only true if x is in the positive domain higher than zero 0 < x .
- In the problem above, the given domain is all real numbers, which include negative numbers. Thus, the statement is False .
"If X=1 ,1<10 and 1> 1/10" True
Relevant wiki: One-step Linear Inequalities
Consider the case x = − 1 , then x 1 = − 1 < 1 0 . However, − 1 < 1 0 1 , hence the answer is f a l s e .