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  # If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Kudos
  Explanation->
  |x|<1 means x should be between -1 and 1.
  
  from (1)
  x2-1<0 => x between -1 and 1. sufficient
  
  from (2)
  if x +ve then same as 1 and sufficient
  if x -ve then |x|=-x
  -x2<1 or x2 > -1 not sufficient
  
  so and A
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  shikhar
  If x≠0, is |x| <1?
  
  (1) x2<1
  (2) |x| < 1/x
  
  Merging similar topics.
  
  shikhar
  Explanation->
  |x|<1 means x should be between -1 and 1.
  
  from (1)
  x2-1<0 => x between -1 and 1. sufficient
  
  from (2)
  if x +ve then same as 1 and sufficient
  if x -ve then |x|=-x
  -x2<1 or x2 > -1 not sufficient
  
  so and A
  
  $$x$$ cannot be negative. Refer to the solution above.
  
  Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross multiply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  boomtangboy
  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  (2) |x| < 1/x
  
  Given question stem asks if |x|<1------> Is -1<x<1
  
  from St 1 we have x^2<1 ------> -1<x<1 So Sufficient
  
  from St2 we have
  
  |x|<1/x
  
  Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation
  
  x= -1/2 so we have 1/2<-2 ------No
  x=3/4 so we have 3/4< 4/3 ------Yes
  x=4/3 so we have 4/3 <3/4 ------- no
  
  We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1
  Ans should be D....
  
  Bunuel's solution is superb. Saves time
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  x<1, x>-1
  -1<x<1
  
  Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
  SUFFICIENT
  
  (2) |x| < 1/x
  Multiply both sides by x
  x^2<1
  Same solution as above.
  SUFFICIENT
  
  (D)
  
  Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  WholeLottaLove
  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  x<1, x>-1
  -1<x<1
  
  Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
  SUFFICIENT
  
  (2) |x| < 1/x
  Multiply both sides by x
  x^2<1
  Same solution as above.
  SUFFICIENT
  
  (D)
  
  Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?
  
  As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.
  
  Hope it helps.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Hi Mahmud,
  
  |x| <1 can be written as -1<x<1 because 1 is a constant BUT
  
  |x| < $$\frac{1}{x}$$ cannot be written as -1/x<x<1/x because 1/x is a variable
  
  Solving,
  
  |x| <$$\frac{1}{x}$$
  
  RHS has to be greater than 0 (As LHS can only be +ve or 0)
  => $$\frac{1}{x}$$ > 0
  => x>0 (x cannot be -ve or 0 ,
  Because, if x is -ve then $$\frac{1}{x}$$ is -ve
  if x = 0 then$$\frac{1}{x}$$ is not defined)
  
  Rgds,
  Rajat
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
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  If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  If $$x\neq{0}$$, is $$|x|$$< 1?
  
  (1) $$x^2$$ < 1
  
  (2) $$|x|$$ < $$1/x$$
  
  If $$x\neq{0}$$, is $$|x|$$< 1?
  This means -1<x<1, where $$x\neq{0}$$
  
  (1) $$x^2$$ < 1
  $$x^2<1$$ means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
  Sufficient
  
  (2) $$|x|$$ < $$1/x$$
  Since |x| >0, we can write 1/|x|x>1
  Since |x| and 1 are positive, x also must be positive for above to be true.
  And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
  Therefore X is between 0 and 1
  Sufficient
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Bunuel
  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
  
  Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.
  
  Thanks.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  mykrasovski
  Bunuel
  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
  
  Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.
  
  Thanks.
  
  Any x, which is between 0 and 1, will obviously fall into the range between -1 and 1, so we can answer the question with an YES: x IS between -1 and 1.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
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  If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Shef08
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
  
  Yes Shef08, you can do this here because $$\frac{1}{x}>|x|$$ means $$\frac{1}{x}\geq{0}$$, which in turn means $$x\geq{0}$$, so you can cross multiply..
  
  exc4libur , as we can see BOTH sides are non-negative, we can square both sides to get our answer ..
  $$\frac{1}{x}>|x|$$......$$\frac{1}{x^2}>|x|^2....x^4<1$$
  Now x>0, so 0<x<1
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  Shef08
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
  
  If we write |x| < 1/x as x < 1/x, then we are assuming that x is positive because |x| = x only when x >= 0.
  
  x < 1/x
  I cannot multiply both sides by x without knowing the sign of x. Here x could be positive or negative e.g.
  when x = 1/2, x < 1/x
  when x = -2, x < 1/x
  
  So I won't multiply both sides by x. I take 1/x to the other side.
  
  x - 1/x < 0
  
  (x^2 - 1)/x < 0
  
  So 0 < x < 1 or x < -1
  But x must be positive so only 0 < x < 1 satisfies.
  
  Now try putting |x| = -x when x is negative.
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  boomtangboy
  If x ≠ 0, is |x| < 1 ?
  
  (1) x^2 < 1
  (2) |x| < 1/x
  
  A Masterpeice from OG
  
  By working on the Question stem, we will have -1<x<1,
  
  S1. x^2 < 1, it means -1<x<1. Sufficient
  S2. |x| < 1/x, it means x has to be positive, because | | is always positive and it is less than a value. So x has to be positive.
  Now as x is positive, and the reciprocal is greater than a number, it means the number has to be between 0 & 1. Sufficient.
  
  Hence D, as both statement alone is sufficient
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  [quote="KarishmaB"]
  
  KarishmaB
  
  Hi Karishma 🙏
  
  (X^2-1)/x<0
  (x-1)(x+1)/x<0
  Now x cannot be zero. As denominator can not zero
  
  But how do we get
  
  0 < x < 1 or x < -1
  
  I understand that x must be positive
  
  Posted from my mobile device
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  puneetfitness
  KarishmaB
  
  KarishmaB
  
  Hi Karishma 🙏
  
  (X^2-1)/x<0
  (x-1)(x+1)/x<0
  Now x cannot be zero. As denominator can not zero
  
  But how do we get
  
  0 < x < 1 or x < -1
  
  I understand that x must be positive
  
  Posted from my mobile device
  
  Check out this post:
  https://gmatclub.com/forum/inequalities ... 91482.html
  
  Let me know if you have any doubts in it.
  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  # If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  1
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  Explanation->
  |x|<1 means x should be between -1 and 1.
  
  from (1)
  x2-1<0 => x between -1 and 1. sufficient
  
  from (2)
  if x +ve then same as 1 and sufficient
  if x -ve then |x|=-x
  -x2<1 or x2 > -1 not sufficient
  
  so and A
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  shikhar
  If x≠0, is |x| <1?
  
  (1) x2<1
  (2) |x| < 1/x
  
  Merging similar topics.
  
  shikhar
  Explanation->
  |x|<1 means x should be between -1 and 1.
  
  from (1)
  x2-1<0 => x between -1 and 1. sufficient
  
  from (2)
  if x +ve then same as 1 and sufficient
  if x -ve then |x|=-x
  -x2<1 or x2 > -1 not sufficient
  
  so and A
  
  $$x$$ cannot be negative. Refer to the solution above.
  
  Also if $$x<0$$ then we have $$-x<\frac{1}{x}$$ and now if we cross multiply by negative $$x$$ then we should flip the sign: $$-x^2>1$$ --> $$x^2+1<0$$ which cannot be true for any real value of $$x$$ (the sum of two positive value cannot be less than zero).
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  boomtangboy
  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  (2) |x| < 1/x
  
  Given question stem asks if |x|<1------> Is -1<x<1
  
  from St 1 we have x^2<1 ------> -1<x<1 So Sufficient
  
  from St2 we have
  
  |x|<1/x
  
  Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation
  
  x= -1/2 so we have 1/2<-2 ------No
  x=3/4 so we have 3/4< 4/3 ------Yes
  x=4/3 so we have 4/3 <3/4 ------- no
  
  We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1
  Ans should be D....
  
  Bunuel's solution is superb. Saves time
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  1
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  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  x<1, x>-1
  -1<x<1
  
  Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
  SUFFICIENT
  
  (2) |x| < 1/x
  Multiply both sides by x
  x^2<1
  Same solution as above.
  SUFFICIENT
  
  (D)
  
  Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  WholeLottaLove
  If x≠0, is |x| < 1 ?
  
  (1) x^2 < 1
  x<1, x>-1
  -1<x<1
  
  Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1.
  SUFFICIENT
  
  (2) |x| < 1/x
  Multiply both sides by x
  x^2<1
  Same solution as above.
  SUFFICIENT
  
  (D)
  
  Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?
  
  As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.
  
  Hope it helps.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Hi Mahmud,
  
  |x| <1 can be written as -1<x<1 because 1 is a constant BUT
  
  |x| < $$\frac{1}{x}$$ cannot be written as -1/x<x<1/x because 1/x is a variable
  
  Solving,
  
  |x| <$$\frac{1}{x}$$
  
  RHS has to be greater than 0 (As LHS can only be +ve or 0)
  => $$\frac{1}{x}$$ > 0
  => x>0 (x cannot be -ve or 0 ,
  Because, if x is -ve then $$\frac{1}{x}$$ is -ve
  if x = 0 then$$\frac{1}{x}$$ is not defined)
  
  Rgds,
  Rajat
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
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  If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  If $$x\neq{0}$$, is $$|x|$$< 1?
  
  (1) $$x^2$$ < 1
  
  (2) $$|x|$$ < $$1/x$$
  
  If $$x\neq{0}$$, is $$|x|$$< 1?
  This means -1<x<1, where $$x\neq{0}$$
  
  (1) $$x^2$$ < 1
  $$x^2<1$$ means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
  Sufficient
  
  (2) $$|x|$$ < $$1/x$$
  Since |x| >0, we can write 1/|x|x>1
  Since |x| and 1 are positive, x also must be positive for above to be true.
  And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
  Therefore X is between 0 and 1
  Sufficient
  
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Bunuel
  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
  
  Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.
  
  Thanks.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  mykrasovski
  Bunuel
  iliavko
  Hi Bunuel!
  
  A question here.
  
  How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?
  
  Do I make myself clear btw?
  
  The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
  
  Can you please elaborate on how a "partial" range in 2) (0 < x < 1) satisfies the full range that is required by the original prompt? I would think that the ranges have to be exact same, i.e. 1) provides a range that is exact same as compared to that in the prompt.
  
  Thanks.
  
  Any x, which is between 0 and 1, will obviously fall into the range between -1 and 1, so we can answer the question with an YES: x IS between -1 and 1.
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
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  If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
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  Shef08
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
  
  Yes Shef08, you can do this here because $$\frac{1}{x}>|x|$$ means $$\frac{1}{x}\geq{0}$$, which in turn means $$x\geq{0}$$, so you can cross multiply..
  
  exc4libur , as we can see BOTH sides are non-negative, we can square both sides to get our answer ..
  $$\frac{1}{x}>|x|$$......$$\frac{1}{x^2}>|x|^2....x^4<1$$
  Now x>0, so 0<x<1
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  Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  Shef08
  This is what I did for statement 2:
  
  |x|<1/x
  As |x| will always be non negative, so the equation becomes:
  X<1/x
  Multiplying both sides by x
  X^2<1
  X^2-1<0 which is then to be worked out as the same way for st 1.
  
  Experts please let me know is this method correct? Bunuel chetan2u VeritasKarishma
  
  Posted from my mobile device
  
  If we write |x| < 1/x as x < 1/x, then we are assuming that x is positive because |x| = x only when x >= 0.
  
  x < 1/x
  I cannot multiply both sides by x without knowing the sign of x. Here x could be positive or negative e.g.
  when x = 1/2, x < 1/x
  when x = -2, x < 1/x
  
  So I won't multiply both sides by x. I take 1/x to the other side.
  
  x - 1/x < 0
  
  (x^2 - 1)/x < 0
  
  So 0 < x < 1 or x < -1
  But x must be positive so only 0 < x < 1 satisfies.
  
  Now try putting |x| = -x when x is negative.
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  boomtangboy
  If x ≠ 0, is |x| < 1 ?
  
  (1) x^2 < 1
  (2) |x| < 1/x
  
  A Masterpeice from OG
  
  By working on the Question stem, we will have -1<x<1,
  
  S1. x^2 < 1, it means -1<x<1. Sufficient
  S2. |x| < 1/x, it means x has to be positive, because | | is always positive and it is less than a value. So x has to be positive.
  Now as x is positive, and the reciprocal is greater than a number, it means the number has to be between 0 & 1. Sufficient.
  
  Hence D, as both statement alone is sufficient
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  [quote="KarishmaB"]
  
  KarishmaB
  
  Hi Karishma 🙏
  
  (X^2-1)/x<0
  (x-1)(x+1)/x<0
  Now x cannot be zero. As denominator can not zero
  
  But how do we get
  
  0 < x < 1 or x < -1
  
  I understand that x must be positive
  
  Posted from my mobile device
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  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  puneetfitness
  KarishmaB
  
  KarishmaB
  
  Hi Karishma 🙏
  
  (X^2-1)/x<0
  (x-1)(x+1)/x<0
  Now x cannot be zero. As denominator can not zero
  
  But how do we get
  
  0 < x < 1 or x < -1
  
  I understand that x must be positive
  
  Posted from my mobile device
  
  Check out this post:
  https://gmatclub.com/forum/inequalities ... 91482.html
  
  Let me know if you have any doubts in it.
  Re: If x 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x [#permalink]
  1   2
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a discussion from a GMAT forum about a specific math problem. The problem is:
  
  **If \( x \neq 0 \), is \( |x| < 1 \)?**
  
  1. \( x^2 < 1 \)
  2. \( |x| < \frac{1}{x} \)
  
  ### Analysis:
  
  1. **Statement 1: \( x^2 < 1 \)**
     - This inequality can be rewritten as \( -1 < x < 1 \).
     - Since \( x \neq 0 \), the valid range for \( x \) is \( -1 < x < 1 \) and \( x \neq 0 \).
     - For any \( x \) in this range, \( |x| < 1 \).
     - Therefore, Statement 1 is sufficient.
  
  2. **Statement 2: \( |x| < \frac{1}{x} \)**
     - Since \( |x| \) is always non-negative, \( \frac{1}{x} \) must also be positive, implying \( x > 0 \).
     - For \( x > 0 \), \( |x| = x \