Let m be an odd integer. Show that the number (m4 + 4m2 + 11)/16 is an integer.

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Let $m$ be an odd integer. Show that the number $\frac{m^4 + 4m^2 + 11}{16}$ is an integer.

Express $m$ as an odd integer.
Let $m = 2k+1$ for some integer $k$ .
Calculate $m^2$ .
$m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$
Since $k$ and $k+1$ are consecutive integers, their product $k(k+1)$ is always an even number. Let $k(k+1) = 2j$ for some integer $j$ .
$m^2 = 4(2j) + 1 = 8j + 1$
Calculate $m^4$ .
$m^4 = (m^2)^2 = (8j+1)^2 = (8j)^2 + 2(8j)(1) + 1^2 = 64j^2 + 16j + 1$
Substitute $m^2$ and $m^4$ into the given expression.
$\frac{m^4 + 4m^2 + 11}{16} = \frac{(64j^2 + 16j + 1) + 4(8j + 1) + 11}{16}$
Simplify the numerator.
$= \frac{64j^2 + 16j + 1 + 32j + 4 + 11}{16}$ $= \frac{64j^2 + (16j + 32j) + (1 + 4 + 11)}{16}$ $= \frac{64j^2 + 48j + 16}{16}$
Divide each term in the numerator by 16.
$= \frac{64j^2}{16} + \frac{48j}{16} + \frac{16}{16}$ $= 4j^2 + 3j + 1$
Since $j$ is an integer, $4j^2$ , $3j$ , and 1 are all integers. The sum of integers is an integer.
Therefore, the expression is an integer.
$\boxed{\text{The expression is an integer.}}$
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If $a, b,$ and $c$ are positive real numbers, then show that: $\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = 1$
Combine the terms under a single square root using the property $\sqrt{x} \times \sqrt{y} \times \sqrt{z} = \sqrt{xyz}$ .
$\sqrt{\frac{b}{a} \times \frac{c}{b} \times \frac{a}{c}}$
Multiply the fractions inside the square root.
$\sqrt{\frac{b \cdot c \cdot a}{a \cdot b \cdot c}}$
Cancel out common terms in the numerator and denominator.
$\sqrt{\frac{abc}{abc}} = \sqrt{1}$
Evaluate the square root.
$\sqrt{1} = 1$
Thus, $\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = 1$ .
$\boxed{\sqrt{\frac{b}{a}} \times \sqrt{\frac{c}{b}} \times \sqrt{\frac{a}{c}} = 1}$
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Suppose $\sqrt{n} = \frac{3}{2}$ , find $-\sqrt{n}$ .
Substitute the given value of $\sqrt{n}$ into the expression $-\sqrt{n}$ .
$-\sqrt{n} = -\left(\frac{3}{2}\right)$
Simplify the expression.
$-\frac{3}{2}$ $\boxed{-\frac{3}{2}}$
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Let $a$ and $b$ be positive real numbers. Show that $\frac{a}{b} + \frac{b}{a} \ge 2$ .
Start with the known inequality that for any real numbers $x$ and $y$ , $(x-y)^2 \ge 0$ .
Expand this inequality:
$x^2 - 2xy + y^2 \ge 0$
Rearrange the terms.
$x^2 + y^2 \ge 2xy$
Let $x = \sqrt{a}$ and $y = \sqrt{b}$ .
Substitute these into the inequality.
$(\sqrt{a})^2 + (\sqrt{b})^2 \ge 2\sqrt{a}\sqrt{b}$ $a + b \ge 2\sqrt{ab}$
This is the AM-GM inequality for $a$ and $b$ .
Step 16

Let m be an odd integer. Show that the number (m4 + 4m2 + 11)/16 is an integer.

Step-by-step solution

:

Express $m$ as an odd integer.

Calculate $m^2$ .

Calculate $m^4$ .

Substitute $m^2$ and $m^4$ into the given expression.

Simplify the numerator.

Divide each term in the numerator by 16.

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Combine the terms under a single square root using the property $\sqrt{x} \times \sqrt{y} \times \sqrt{z} = \sqrt{xyz}$ .

Multiply the fractions inside the square root.

Cancel out common terms in the numerator and denominator.

Evaluate the square root.

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Substitute the given value of $\sqrt{n}$ into the expression $-\sqrt{n}$ .

Simplify the expression.

:

Start with the known inequality that for any real numbers $x$ and $y$ , $(x-y)^2 \ge 0$ .

Rearrange the terms.

Let $x = \sqrt{a}$ and $y = \sqrt{b}$ .

Step 16

Still stuck on this one?

More Mathematics Questions

Let m be an odd integer. Show that the number (m4 + 4m2 + 11)/16 is an integer.

Step-by-step solution

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Express mmm as an odd integer.

Calculate m2m^2m2.

Calculate m4m^4m4.

Substitute m2m^2m2 and m4m^4m4 into the given expression.

Simplify the numerator.

Divide each term in the numerator by 16.

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Combine the terms under a single square root using the property x×y×z=xyz\sqrt{x} \times \sqrt{y} \times \sqrt{z} = \sqrt{xyz}x​×y​×z​=xyz​.

Multiply the fractions inside the square root.

Cancel out common terms in the numerator and denominator.

Evaluate the square root.

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Substitute the given value of n\sqrt{n}n​ into the expression −n-\sqrt{n}−n​.

Simplify the expression.

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Start with the known inequality that for any real numbers xxx and yyy, (x−y)2≥0(x-y)^2 \ge 0(x−y)2≥0.

Rearrange the terms.

Let x=ax = \sqrt{a}x=a​ and y=by = \sqrt{b}y=b​.

Step 16

Still stuck on this one?

More Mathematics Questions

Express $m$ as an odd integer.

Calculate $m^2$ .

Calculate $m^4$ .

Substitute $m^2$ and $m^4$ into the given expression.

Combine the terms under a single square root using the property $\sqrt{x} \times \sqrt{y} \times \sqrt{z} = \sqrt{xyz}$ .

Substitute the given value of $\sqrt{n}$ into the expression $-\sqrt{n}$ .

Start with the known inequality that for any real numbers $x$ and $y$ , $(x-y)^2 \ge 0$ .

Let $x = \sqrt{a}$ and $y = \sqrt{b}$ .