Multiplicative Inverse Proof Problem

[A.Magnus]

I am working on a proof problem on ordered field from a textbook, which lists additive and multiplicative properties similar to the ones here:

If $x \ne 0$, show that $\frac{1}{\frac{1}{x}} = x.$

The followings are what I was able to come out -- I just wanted to make sure that they are acceptable:

(a) By the multiplicative inverse property, there must exists $\frac{1}{x}$ such that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = 1$.

(b) By the same property, there must exists $x$ such that $\frac{1}{x} \cdot x = 1$.

(c) By equating the (a) and (b) above, we have $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = \frac{1}{x} \cdot x$.

(d) By multiplying both sides with $x$, then we have $x \cdot \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = x \cdot \frac{1}{x} \cdot x$, showing that $\frac{1}{\frac{1}{x}} = x$, as desired.

Thank you for your time and gracious helps. ~MA

 

[I like Serena]

I am working on a proof problem on ordered field from a textbook, which lists additive and multiplicative properties similar to the ones here:

The followings are what I was able to come out -- I just wanted to make sure that they are acceptable:

(a) By the multiplicative inverse property, there must exists $\frac{1}{x}$ such that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = 1$.

Hey again MaryAnn,

This is not quite correct.
Instead we have that given $x\ne 0$ by the multiplicative inverse property, there must exist an $\frac{1}{x}$ such that $\frac{1}{x} \cdot x = 1$.

Now we know that $\frac{1}{x}$, but we do not know if it is equal to $0$ or not.
Therefore we cannot deduce yet either that its multiplicative inverse exists, can we?
It's only as soon as we can conclude that $\frac{1}{x}\ne 0$, that we can conclude by the multiplicate inverse property that $\frac{1}{\frac{1}{x}}$ must exist, and that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = 1$.

(b) By the same property, there must exists $x$ such that $\frac{1}{x} \cdot x = 1$.

This is the wrong way around. We conclude from the existence of $x\ne 0$ that $\frac 1x$ exists.

(c) By equating the (a) and (b) above, we have $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = \frac{1}{x} \cdot x$.

This is a correct deduction.

(d) By multiplying both sides with $x$, then we have $x \cdot \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = x \cdot \frac{1}{x} \cdot x$, showing that $\frac{1}{\frac{1}{x}} = x$, as desired.

This is true, but you are skipping a couple of steps, and you are not mentioning which properties you are using.
As it is, you are using the property of associativity (twice), the property of the multiplicative inverse (twice), and the property of unity (twice).

Thank you for your time and gracious helps. ~MA

 

[A.Magnus]

Thank you for your response! ~MA

 

[A.Magnus]

How about these revised steps - hope these come out right:

(a) Since $x \neq 0$, there must exist $\frac{1}{x}$ such that $x \cdot \frac{1}{x} = 1$. (Multiplicative Inverse Property)

(b) By Commutative Property, we have $\frac{1}{x} \cdot x = 1$.

(c) Multiplying both sides of the equation by $\frac{1}{\frac{1}{x}}$, we have $\frac{1}{\frac{1}{x}} \cdot \frac{1}{x} \cdot x = 1 \cdot \frac{1}{\frac{1}{x}}$.

(d) By Property of Unity, in the right hand side we have $\frac{1}{\frac{1}{x}} \cdot \frac{1}{x} \cdot x = \frac{1}{\frac{1}{x}}$.

(e) By Associative Property, we then have $(\frac{1}{\frac{1}{x}} \cdot \frac{1}{x}) \cdot x = \frac{1}{\frac{1}{x}}$.

(f) By Property of Unity, we finally have $1 \cdot x = \frac{1}{\frac{1}{x}}$, and by the same property again $x = \frac{1}{\frac{1}{x}}$, as desired.

Thank you again for all your helps. ~MA

 

[I like Serena]

How about these revised steps - hope these come out right:

(a) Since $x \neq 0$, there must exist $\frac{1}{x}$ such that $x \cdot \frac{1}{x} = 1$. (Multiplicative Inverse Property)

(b) By Commutative Property, we have $\frac{1}{x} \cdot x = 1$.

(c) Multiplying both sides of the equation by $\frac{1}{\frac{1}{x}}$, we have $\frac{1}{\frac{1}{x}} \cdot \frac{1}{x} \cdot x = 1 \cdot \frac{1}{\frac{1}{x}}$.

Careful, we should multiply on the same side, since we should not assume commutativity of multiplication.
And we should not assume associativity by leaving out the parentheses.
So it should be: $\frac{1}{\frac{1}{x}} \cdot \left(\frac{1}{x} \cdot x\right) = \frac{1}{\frac{1}{x}} \cdot 1$

(d) By Property of Unity, in the right hand side we have $\frac{1}{\frac{1}{x}} \cdot \frac{1}{x} \cdot x = \frac{1}{\frac{1}{x}}$.

(e) By Associative Property, we then have $(\frac{1}{\frac{1}{x}} \cdot \frac{1}{x}) \cdot x = \frac{1}{\frac{1}{x}}$.

(f) By Property of Unity, we finally have $1 \cdot x = \frac{1}{\frac{1}{x}}$, and by the same property again $x = \frac{1}{\frac{1}{x}}$, as desired.

That's the property of multiplicative inverse instead of unity.
The second part is using the property of unity though.