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# Find the truth set in case of each of the following open sentences defined on N:

Question:

Find the truth set in case of each of the following open sentences defined on N:

(i) $x+2<10$

(ii) $x+5<4$

(iii) $x+3>2$

Solution:

The open sentence x + 2 < 10 is defined on N; the set of natural numbers.

$\mathrm{N}:\{1,2,3,4 \ldots\}$

$x=1 \rightarrow x+2=3<10$

$x=2 \rightarrow x+2=4<10$

$x=3 \rightarrow x+2=5<10$

$x=4 \rightarrow x+2=6<10$

$x=5 \rightarrow x+2=7<10$

$x=6 \rightarrow x+2=8<10$

$x=7 \rightarrow x+2=9<10$

$x=8 \rightarrow x+2=10$

So, $\exists x \in N$, such that $x+2<10$

$x=\{1,2,3,4,5,6,7\}$ satisfies $x+2<10$.

So, the truth set of open sentence $x+2<10$ defined on $N$ is, $\{1,2,3,4,5,6,7\}$

(ii) The open sentence $x+5<4$ is defined on $N$; the set of natural numbers.

$N:\{1,2,3,4 \ldots\}$

$x=1 \rightarrow 1+5=6>4$

So, the truth set of open sentence $x+5<4$ defined on $N$ is an empty set, \{\} .

(iii) The open sentence $x+3>2$ is defined on $N$; the set of natural numbers.

$\mathrm{N}:\{1,2,3,4 \ldots\}$

$x=1 \rightarrow x+3=4>2$

$x=2 \rightarrow x+3=5>2$

$x=3 \rightarrow x+3=6>2$

$x=4 \rightarrow x+3=7>2$

$x=5 \rightarrow x+3=8>2$

$x=6 \rightarrow x+3=9>2$

And so on

So, $\exists x \in N$, such that $x+3>2$

$x=\{1,2,3,4,5,6,7 \ldots\}$ satisfies $x+3>2$.

So, the truth set of open sentence $x+3>2$ defined on $N$ is an infinite set as there is infinite natural numbers satisfying the equation $x+3>2$.

$\{1,2,3,4,5,6,7 \ldots\}$

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