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umka2103 [35]
3 years ago
15

Prove that if a and b are integers, then a^2-4b egal or non-egal 2

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
7 0

Answer:

tex]a^2 - 4b \neq 2[/tex]

Step-by-step explanation:

We are given that a and b are integers, then we need to show that a^2 - 4b \neq 2

Let  a^2 - 4b = 2

If a is an even integer, then it can be written as a = 2c, then,

a^2 - 4b = 2\\(2c)^2 - 4b =2\\4(c^2 -b) = 2\\(c^2 -b) =\frac{1}{2}

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

If a is an odd integer, then it can be written as a = 2c+1, then,

a^2 - 4b = 2\\(2c+1)^2 - 4b =2\\4(c^2+c-b) = 2\\(c^2+c-b) =\frac{1}{4}

RHS is a fraction but LHS can never be a fraction, thus it is impossible.

Thus, our assumption was wrong and a^2 - 4b \neq 2.

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Answer:

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Different ways to find the product:

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3 years ago
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