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ajprincess

  • 2 years ago

Please help:) Prove that if a|b and b|a, then \(a=\pm b\)

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  1. sauravshakya
    • 2 years ago
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    Let x=a/b then since b divides a, x is a integer other than 0 Now, Since a also divides b b/a is also integer other than 0 Here, b/a =1/(a/b) =1/x So, x can be either +1 or -1

  2. sauravshakya
    • 2 years ago
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    Now, when x=1 then a/b=1 ===>a=b when x=-1 then a/b=-1 =====>a=-b

  3. sauravshakya
    • 2 years ago
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    Thus, a=+-b

  4. ajprincess
    • 2 years ago
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    Thanx a lottt @sauravshakya.:)

  5. jiteshmeghwal9
    • 2 years ago
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    @sauravshakya i couldn't understand ur statement ``` Now, Since a also divides b b/a is also integer other than 0 ```

  6. jiteshmeghwal9
    • 2 years ago
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    @ajprincess can u tell me the statement in easier way ??

  7. ajprincess
    • 2 years ago
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    a|b is a notation that means a divides b and b/a is an integer. b|a is a notation that means b divides a and a/b is an integer.

  8. ajprincess
    • 2 years ago
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    a|b can also be expressed in this way b/a=x, x is an integer. If x is o , b must be equal to 0. b|a can also be expressed in this way a/b=y, y is an integer a/0=infinity. infinity is not an ineger that is y we give the condition that x is an integer except 0. getting it @jiteshmeghwal9?

  9. ParthKohli
    • 2 years ago
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    Hmm, I take it that you meant a/0 is not defined.

  10. jiteshmeghwal9
    • 2 years ago
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    ya! I gt it a little but \(\frac{a}{b}\) doesn't this mean b divides a

  11. ajprincess
    • 2 years ago
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    ya u can take it that way @parthkohli

  12. ajprincess
    • 2 years ago
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    a|b and a/b are different.

  13. ParthKohli
    • 2 years ago
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    \[a\ne 0, b \ne 0\]because if one of them is zero, then one of \(a|b\) or \(b|a\) will be false. =)

  14. jiteshmeghwal9
    • 2 years ago
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    ohh ! i thought it is a/b i gt it completely, thanx :)

  15. ParthKohli
    • 2 years ago
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    \(b|a\) is the same as \(bx = a :x\in \mathbb{Z}^{+}\)

  16. ajprincess
    • 2 years ago
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    ya u are absolutely right:) @ParthKohli and welcome:) @jiteshmeghwal9

  17. ParthKohli
    • 2 years ago
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    Nice proof! @sauravshakya

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