I'm new to proofs. Would someone please give me an opinion on my proof?

[nietzsche]

My prof gave us twelve basic properties of numbers, and I think I'm supposed to use those in my proof, but I'm not sure how to incorporate them.

The properties are:
P1 Associative law for addition
P2 Additive identity
P3 Additive inverse
P4 Commutative law for addition
P5 Associative law for multiplication
P6 Multiplicative identity
P7 Multiplicative inverse
P8 Commutative law for multiplication
P9 Distributive law
P10 Trichotomy law
P11 Closure under addition
P12 Closure under multiplication

Homework Statement

Prove that if 0 < a < b, then

a &lt; \sqrt{ab} &lt; \frac{a + b}{2} &lt; b

Homework Equations

N/A

The Attempt at a Solution

Part I:

<br /> \begin{align*}<br /> a&amp;&lt;b\\<br /> a^2&amp;&lt;ab\\<br /> \sqrt{a^2}&amp;&lt;\sqrt{ab}\\<br /> a&amp;&lt;\sqrt{ab}<br /> \end{align*}<br />

Part II:

<br /> Suppose:<br /> \begin{align*}<br /> \sqrt{ab}&amp;\geq\frac{a+b}{2}\\<br /> ab&amp;\geq\left(\frac{a+b}{2}\right)^2\\<br /> ab&amp;\geq\frac{a^2+2ab+b^2}{4}\\<br /> 4ab&amp;\geq a^2+2ab+b^2\\<br /> 0&amp;\geq a^2-2ab+b^2\\<br /> 0&amp;\geq (a-b)^2\\<br /> \end{align*}<br />

<br /> but:<br /> \begin{align*}<br /> 0&amp;&lt;(a-b)^2\\<br /> \therefore \sqrt{ab}&amp;&lt;\frac{a+b}{2}<br /> \end{align*}<br />

Part III:

<br /> \begin{align*}<br /> a&amp;&lt;b\\<br /> a+b &amp;&lt; b+b\\<br /> a+b &amp;&lt; 2b\\<br /> \frac{a+b}{2} &amp;&lt; b<br /> \end{align*}<br />

So I'm able to prove them, but I don't know if I used the properties correctly (if at all). Any opinions or suggestions?

 

[Mark44]

My prof gave us twelve basic properties of numbers, and I think I'm supposed to use those in my proof, but I'm not sure how to incorporate them.

The properties are:
P1 Associative law for addition
P2 Additive identity
P3 Additive inverse
P4 Commutative law for addition
P5 Associative law for multiplication
P6 Multiplicative identity
P7 Multiplicative inverse
P8 Commutative law for multiplication
P9 Distributive law
P10 Trichotomy law
P11 Closure under addition
P12 Closure under multiplication

Homework Statement

Prove that if 0 < a < b, then

a &lt; \sqrt{ab} &lt; \frac{a + b}{2} &lt; b

Homework Equations

N/A

The Attempt at a Solution

Part I:

<br /> \begin{align*}<br /> a&amp;&lt;b\\<br /> a^2&amp;&lt;ab\\<br /> \sqrt{a^2}&amp;&lt;\sqrt{ab}\\<br /> a&amp;&lt;\sqrt{ab}<br /> \end{align*}<br />

By "using the properties" I think your prof means for you to indicate which property allows you to do each step. However, in some of your steps you are using operations that aren't listed amongst the properties you show. For example, in your 2nd inequality, when you multiply both members of an inequality by a positive number, the direction of the inequality stays the same.

Part II:

<br /> Suppose:<br /> \begin{align*}<br /> \sqrt{ab}&amp;\geq\frac{a+b}{2}\\<br /> ab&amp;\geq\left(\frac{a+b}{2}\right)^2\\<br /> ab&amp;\geq\frac{a^2+2ab+b^2}{4}\\<br /> 4ab&amp;\geq a^2+2ab+b^2\\<br /> 0&amp;\geq a^2-2ab+b^2\\<br /> 0&amp;\geq (a-b)^2\\<br /> \end{align*}<br />

<br /> but:<br /> \begin{align*}<br /> 0&amp;&lt;(a-b)^2\\<br /> \therefore \sqrt{ab}&amp;&lt;\frac{a+b}{2}<br /> \end{align*}<br />

For the one above, instead of doing a proof by contradiction, as you have done, it would be simpler to start with (a - b)² >= 0 (the square of any real number is always nonnegative). Then expand the left side and you should be able to get to the conclusion you need.

Part III:

<br /> \begin{align*}<br /> a&amp;&lt;b\\<br /> a+b &amp;&lt; b+b\\<br /> a+b &amp;&lt; 2b\\<br /> \frac{a+b}{2} &amp;&lt; b<br /> \end{align*}<br />

So I'm able to prove them, but I don't know if I used the properties correctly (if at all). Any opinions or suggestions?

In the one above, b + b = b(1 + 1) = b*2 = 2b. The properties used are the distributive property and the commutative property of multiplication.