Respuesta :
Answer:
a) True
b) True
c) False
d) True
e) False
f) True
g) False
Step-by-step explanation:
a) Q(0) is true because;
put x= 0 in x+1>2x
==> 0+1>2×0
==> 1>0
b) Q(-1) is true because;
if we put x= -1 in x+1>2x
==> -1+1>2 (-1)
==> 0> -2
c) Q(1) is false because;
if we put x=1 in x+1>2x
==> 1+1> 2 ×1
==> 2>2
which is not true!
d) the statement is true because;
If we put x=0 in x+1>2x
==> 0+1>2×0
==> 1>0
e) The statement is false because;
if we put x=1 then statement becomes false as well !
f) The statement is true because;
If x=3
3+1≤2×3
==> 4≤6
So our statement is true
g) The statement is false because
If we suppose x=0 and x+1≤2x
then,
0+1≤2×0
==> 1≤0
Which is false - so x+1≥2x is true
Inequalities help us to compare two unequal expressions.
What are inequalities?
Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed.
It is mostly denoted by the symbol <, >, ≤, and ≥.
In order to know the truth values for the statement, we need to substitute the values in the given inequality, (x+1>2x), therefore,
a) Q(0)
If we substitute the value of x as 0, then the statement is true because, if we put x=0 in [tex]x+1>2x[/tex], then
[tex]x+1>2x\\(0) + 1 > 2(0)\\1 > 0[/tex]
Hence, the statement Q(x) holds true for the value of x as 0.
b) Q(-1)
If we substitute the value of x as -1, then the statement is true because, if we put x=-1 in [tex]x+1>2x[/tex], then
[tex]x+1>2x\\(-1) + 1 > 2(-1)\\0 > -2[/tex]
Hence, the statement Q(x) holds true for the value of x as -1.
c) Q(1) is false because;
If we substitute the value of x as 1, then the statement becomes false because, if we put x=1 in [tex]x+1>2x[/tex], then
[tex]x+1>2x\\(1) + 1 > 2(1)\\2 > 2[/tex]
Hence, the statement Q(x) is false for the value of x as 1.
d) ∃xQ(x) There exists an x such that Q(x)
If we substitute the value of x as 0, then the statement is true because, if we put x=0 in [tex]x+1>2x[/tex], then
[tex]x+1>2x\\(0) + 1 > 2(0)\\1 > 0[/tex]
Hence, the statement Q(x) holds true ∃xQ(x) There exists an x such that Q(x).
e). ∀xQ(x) For all x such that Q(x)
If we substitute the value of x as 1, then the statement becomes false because, if we put x=1 in [tex]x+1>2x[/tex], then
[tex]x+1>2x\\(1) + 1 > 2(1)\\2 > 2[/tex]
Hence, the statement Q(x) is false for ∀xQ(x) For all x such that Q(x).
f) ∃x¬Q(x)
If we substitute the value of x as 3, then the statement is true because, if we put x=3 in [tex]x+1\leq 2x[/tex], then
[tex]x+1\leq 2x\\(5) + 1 \leq 2(5)\\6 \leq 10[/tex]
Hence, the statement Q(x) holds true for ∃x¬Q(x).
g) ∀x¬Q(x)
If we substitute the value of x as 0, then the statement is false because, if we put x=0 in [tex]x+1\leq 2x[/tex], then
[tex]x+1\leq 2x\\(0) + 1 \leq 2(0)\\1 \leq 0[/tex]
Hence, the statement Q(x) is true for ∀x¬Q(x).
Learn more about Inequalities:
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