Login/Sign Up
Home » Show that the statement p: “If x is a real number such that x^3 + 4x = 0, then x is 0” is true by method of contrapositive
Show that the statement p: “If x is a real number such that x^3 + 4x = 0, then x is 0” is true by method of contrapositive
CBSE | Class 11 | Exercise 14.5 | Mathematical Reasoning | Maths | NCERT
Show that the statement p: “If x is a real number such that x^3 + 4x = 0, then x is 0” is true by method of contrapositive

Let

    \[p:\]

‘In case

    \[x\]

is a genuine number to such an extent that

    \[{{x}^{3}}\text{ }+\text{ }4x\text{ }=\text{ }0,\]

then, at that point,

    \[x\text{ }is\text{ }0'\]

    \[q:\text{ }x\]

is a genuine number to such an extent that

    \[{{x}^{3}}\text{ }+\text{ }4x\text{ }=\text{ }0\]

    \[r:\text{ }x\text{ }is\text{ }0\]

By contrapositive strategy, to show proclamation

    \[p\]

to be valid, we expect that

    \[r\]

is bogus and prove that

    \[q\]

should be bougus

    \[\sim r:~x~\ne \text{ }0\]

Unmistakably, it tends to be seen that

    \[({{x}^{2}}~+\text{ }4)~\]

will consistently be positive

    \[x\text{ }\ne \text{ }0\]

infers that the result of any sure genuine number with

    \[x\]

isn’t zero.

Presently, consider the result of

    \[x\]

with

    \[({{x}^{2}}~+\text{ }4)\]

    \[\begin{array}{*{35}{l}} \therefore ~x~\left( {{x}^{2}}~+\text{ }4 \right)\text{ }\ne \text{ }0 \\ {{x}^{3}}~+\text{ }4x\text{ }\ne \text{ }0 \\ \end{array}\]

This shows that assertion

    \[q\]

isn’t correct.

Thus, demonstrated that

    \[\sim r\Rightarrow \sim q\]

Thus, the given assertion

    \[p\]

is valid.

i

More Material