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3 years ago

14

what is the solution to the equation fraction 1 over 3(n − fraction 1 over 3) = fraction 1 over 6n? n = 3 n = 2 n = fraction 5 o

ver 6 n = fraction 2 over 3

1 answer:

Nat2105 [25]3 years ago

3 0

1/3(n - 1/3) = 1/6n
1/3n - 1/9 = 1/6n
-1/9 = 1/6n - 1/3n
-1/9 = 1/6n - 2/6n
-1/9 = - 1/6n
-1/9 * -6 = n
6/9 = n
2/3 = n

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Problems 2.21, 2.22, 2.23

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Statements can be proved by contrapositive, contradiction or by induction.

<em>2.21 and 2.23 are proved by contrapositive</em>
<em>2.22 is proved by induction</em>

<u />

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<u>2.21: If </u> $n^3$ <u> is even, then n is even (By contrapositive)</u>

The contrapositive of the above statement is that:

<em>If n is odd, then </em> $n^3$ <em> is odd</em>

Represent the value of n as:

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Group

$n^3 =[ 8k^3 + 6k^2 + 6k] + 1$

Factor out 2

$n^3 =2[4k^3 + 3k^2 + 3k] + 1$

Assume w is an integer; where:

$w =4k^3 + 3k^2 + 3k$

So, we have:

$n^3 =2w + 1$

The constant term (i.e. 1) means that $n^3$ is odd.

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<u>2.22 </u> $3n + 4$ <u> is even, if and only if n is even</u>

We have: $3n + 4$ <u />

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$3n + 4 = 3(2k + 2) + 4$

Open bracket

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$3n + 4 = 2(3k + 5)$

The factor of 2 means that $3n + 4$ is even.

<em>Hence, </em> $3n + 4$ <em> is even, if and only if n is even </em>

<em />

<u />

<u>2.22: </u> $s \ne -1$ <u> and </u> $t \ne -1$ <u>, then </u> $s + t + st \ne -1$ <u />

To do this, we prove by contrapositive.

The contrapositive of the above statement is:

If $s = -1$ and $t=-1$ , then $s + t + st = -1$

We have:

$s + t + st = -1$

Substitute the values of s and t in: $s + t + st = -1$

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If $s = -1$ and $t=-1$ , then $s + t + st = -1$

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