Respuesta :
The statement above establishes the following:
If:
(1) [tex]a=b[/tex]
(2) [tex]c \neq 0[/tex]
Then we must assume that in fact [tex]a=b[/tex], therefore if we multiply the same number, say c, to both sides of an equation this one is not affected, then we have:
(3) [tex]ac=bc \rightarrow c(a-b)=0[/tex]
First answer.
Through division, the c's cancel and zero. Let's prove it:
a. If we divide (3) by c, this is canceled and the equation (3) is converted to:
[tex]a-b=0[/tex] All right up here
b. If we divide (3) by (a - b), this term is canceled and the equation (3) is converted to:
[tex]c=0[/tex] But this is a contradiction given that in the statement above [tex]c \neq 0[/tex]
Second answer. Which statement accurately completes the proof?
The statement that completes the proof is necessarily:
[tex]a \neq b [/tex]
For example:
[tex]a=5[/tex]
[tex]b=4[/tex]
[tex]c=9[/tex]
Given that:
[tex]a \neq b[/tex] and [tex]c \neq 0[/tex]
Then:
[tex]ac \neq bc \rightarrow 5\times9 \neq 4\times9 \rightarrow 45 \neq 36[/tex]
If:
(1) [tex]a=b[/tex]
(2) [tex]c \neq 0[/tex]
Then we must assume that in fact [tex]a=b[/tex], therefore if we multiply the same number, say c, to both sides of an equation this one is not affected, then we have:
(3) [tex]ac=bc \rightarrow c(a-b)=0[/tex]
First answer.
Through division, the c's cancel and zero. Let's prove it:
a. If we divide (3) by c, this is canceled and the equation (3) is converted to:
[tex]a-b=0[/tex] All right up here
b. If we divide (3) by (a - b), this term is canceled and the equation (3) is converted to:
[tex]c=0[/tex] But this is a contradiction given that in the statement above [tex]c \neq 0[/tex]
Second answer. Which statement accurately completes the proof?
The statement that completes the proof is necessarily:
[tex]a \neq b [/tex]
For example:
[tex]a=5[/tex]
[tex]b=4[/tex]
[tex]c=9[/tex]
Given that:
[tex]a \neq b[/tex] and [tex]c \neq 0[/tex]
Then:
[tex]ac \neq bc \rightarrow 5\times9 \neq 4\times9 \rightarrow 45 \neq 36[/tex]