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Show that the statement p: If x is a real number such that `x^3+4x=0`. then x is 0 is true by(i) direct method, (ii) method of contradiction,(iii) method of contrapositive. |
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Answer» Given statement : if x is a real number such that `x^3+4x=0` , then ` x=0` Comonent p:x is a real number such that `x^3+4x=0` Component `q:x=0` (i) Direct method : `because x ` is a real number and `x^3+4x=0` `therefore x(x^2+4)=0` `rArrx=0` or ` x^2+4=0` But `x^2+4` cannot be zero. `therefore x=0` (ii) Method of contradication : Let x is a real number and `x^3+4x=0` let `xne0` Now `x(x^2+4)=0` `rArr(x^2+4)=0` `rArrx=-` or `x^2+4=0` But `x^2+4` cannot be zero. `therefore x=0` (iii) Method of contrapositive : let `xne0` `rArrx` is a real number and `xne0` `therefore x^2gt0` `rArrx^2+4gt0` `becausexne0` and `x^2+4gt0` `rArrxne0` and `x^2+4ne0` `rArrx(x^2+4)ne0` `rArrx^3+4xne0` therefore, `xne 0 rArrx^3+4xne0` `rArrx^3+4x=0rArr x=0` |
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