Respuesta :
Step-by-step explanation:
Proposition If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex], then [tex]a^{2}-4b \neq2[/tex]
You can prove this proposition by contradiction, you assume that the statement is not true, and then show that the consequences of this are not possible.
Suppose the proposition If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex], then [tex]a^{2}-4b \neq2[/tex] is false. Thus there exist integers If a, b [tex]\in[/tex] [tex]\mathbb{Z}[/tex] for which [tex]a^{2}-4b=2[/tex]
From this equation you get [tex]a^{2}=4b+2=2(2b+1)[/tex] so [tex]a^{2}[/tex] is even. Since [tex]a^{2}[/tex] is even, a is even, this means [tex]a=2d[/tex] for some integer d. Next put [tex]a=2d[/tex] into [tex]a^{2}-4b=2[/tex]. You get [tex] (2d)^{2}-4b=2[/tex] so [tex]4(d)^{2}-4b=2[/tex]. Dividing by 2, you get [tex]2(d)^{2}-2b=1[/tex]. Therefore [tex]2((d)^{2}-b)=1 [/tex], and since [tex](d)^{2}-b[/tex] [tex]\in[/tex] [tex]\mathbb{Z}[/tex], it follows that 1 is even.
And that is the contradiction because 1 is not even. In other words, we were wrong to assume the proposition was false. Thus the proposition is true.
