Prove that if n is a perfect square, then n+1 can never be a perfect square

Mathematics

1 answer:

natka813 [3]3 years ago

8 0

Answer:

Proved

Step-by-step explanation:

To prove that if n is a perfect square, then n+1 can never be a perfect square

Let n be a perfect square

$n=x^2$

Let $n+1 = y^2$

Subtract to get

$1 = y^2-x^2 =(y+x)(y-x)$

Solution is y+x=y-x=1

This gives x=0

So only 0 and 1 are consecutive integers which are perfect squares

No other integer satisfies y+x=y-x=1

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Question 1:
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First, apply the common denominator to the first fraction:
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Do the same for the second: $15( \frac{5}{7}) \\ \frac{5}{7}* \frac{15}{15} \\ \frac{5*15}{7*15} \\ \frac{75}{105}$
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Question Two:
$\frac{5}{9}$ ? $\frac{7}{13}$
Apply the common denominator to fraction one:
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Fraction two:
$9(\frac{7}{13}) \\ \frac{7}{13} * \frac{9}{9} \\ \frac{7*9}{13*9} \\ \frac{63}{117}$
Evaluate:
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