Proof by Induction: Solve Complex Variables Problem 12

In summary, when using mathematical induction to prove a statement is true, one must first show it is true for the lowest value and then assume it holds true for some value k. Then, it must be shown that if it is true for the value (k+1), it is also true for all values. In the given example, the statement is true for n = 1, but to properly prove it, the induction should start with n = 2.
  • #1
Tsunoyukami
215
11
I'm having a little bit of difficulty with proofs by induction. I believe I understand the principles behind induction but find that I often get "stuck" in my proof - and I can "see" that its true but am not sure how to get that one step that will finish the proof.

When using mathematical induction to show that a statement is true we first show that it is true for the lowest value (usually 0 or 1) and then assume it holds true for some value k > 0 (or 1). Then we show that if the statement is true for the value (k+1) that it is true for all values, correct?So here is a question I must complete for homework from Stephen D. Fisher's Complex Variables, 2e:

"12. Let ##z_1, z_2, ..., z_n## be complex numbers. Establish the following formulas by mathematical induction:

a) ##|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|##" (Section 1.1, page 9).
Here is my attempt at a solution:First we show that this statement is true for n = 1 (which is obvious) - we find:

##|z_1| = |z_1|##

Next we assume that this statement holds true for some k > 1; that is that the following is true:

##|z_1 z_2 ... z_k| = |z_1| |z_2| ... |z_k|##

Lastly we must show that for some value, (k+1), the statement ##|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|## is true. So:

##|z_1 z_2 ... z_k z_{k+1}| = |z_1| |z_2| ... |z_k| |z_{k+1}| (1)##

To me this seems obvious is I can write:

##|z_1 z_2 ... z_k z_{k+1}| = |z_1 z_2 ... z_k| |z_{k+1}| (2)##
##|z_1 z_2 ... z_k| |z_{k+1}| = |z_1| |z_2| ... |z_k| |z_{k+1}| (3)##

By using the assumption that it holds true for k > 1. Is this a valid step to make? Can I go from (1) to (2), from which point (3) readily follows? Or am I missing something?

Any assistance is much appreciated; please provide me confirmation or a hint to lead me in the right direction - thanks!
 
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  • #2
You need to prove that [itex]|z_1z_2| = |z_1||z_2|[/itex] then apply the inductive hypothesis.
 
  • #3
Tsunoyukami said:
I'm having a little bit of difficulty with proofs by induction. I believe I understand the principles behind induction but find that I often get "stuck" in my proof - and I can "see" that its true but am not sure how to get that one step that will finish the proof.

When using mathematical induction to show that a statement is true we first show that it is true for the lowest value (usually 0 or 1) and then assume it holds true for some value k > 0 (or 1). Then we show that if the statement is true for the value (k+1) that it is true for all values, correct?So here is a question I must complete for homework from Stephen D. Fisher's Complex Variables, 2e:

"12. Let ##z_1, z_2, ..., z_n## be complex numbers. Establish the following formulas by mathematical induction:

a) ##|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|##" (Section 1.1, page 9).
Here is my attempt at a solution:First we show that this statement is true for n = 1 (which is obvious) - we find:

##|z_1| = |z_1|##
You should start with n = 2. I will explain why below.
Next we assume that this statement holds true for some k > 2; that is that the following is true:

##|z_1 z_2 ... z_k| = |z_1| |z_2| ... |z_k|##

Lastly we must show that for some value, (k+1), the statement ##|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|## is true.

No, we don't show it is true for "some value". We show that if it is true for ##n=k## it is true the next value, which is ##k+1##
So:

##|z_1 z_2 ... z_k z_{k+1}| = |z_1| |z_2| ... |z_k| |z_{k+1}| (1)##

To me this seems obvious is I can write:

##|z_1 z_2 ... z_k z_{k+1}| = |z_1 z_2 ... z_k| |z_{k+1}| (2)##

That assumes you know it works for 2 complex numbers, which you haven't proven.

##|z_1 z_2 ... z_k| |z_{k+1}| = |z_1| |z_2| ... |z_k| |z_{k+1}| (3)##

By using the assumption that it holds true for k > 1. Is this a valid step to make? Can I go from (1) to (2), from which point (3) readily follows? Or am I missing something?

Any assistance is much appreciated; please provide me confirmation or a hint to lead me in the right direction - thanks!
That you can go from step 2 to step 3 is because your induction hypothesis is that it works for ##n=k##. But notice that this argument doesn't work for ##k=2##. That is why you need to start with ##k=2##.

There is a common example to illustrate the fallacy your argument falls into.

Theorem. If one man in a group of n men is a millionaire, they all are millionaires.

Proof: n=1 Obvious
Assume it is true for n = k, that is, if one man in a group of k men is a millionaire, they all are.

Now suppose you have a group of k+1 men, and one of them is a millionaire. Divide them into two groups, one containing k men and the other 1 man, with the millionaire in the larger group. By the induction hypothesis, 1 man in the group of k men is a millionaire so all k of that group are. Now take 1 of the millionaires out of the larger group and exchange him with the lone man. Now by the same argument, the man who was the lone man is also a millionaire, so everyone is.
 
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  • #4
Okay...so using ##z_1 = x_1 + iy_1## and ##z_2 = x_2 + iy_2##:

##|z_1 z_2| = |(x_1 + iy_1)(x_2 +iy_2)| = |(x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)|##

Now using definition of ##|z|## is such that ##|z| = |x + iy| = \sqrt{x^2 + y^2}## we find:

##|z_1 z_2| = |(x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)| = \sqrt{(x_1 x_2 - y_1 y_2)^2 + (x_1 y_2 + x_2 y_1)^2}##

Which, when simplified, yields:

##|z_1 z_2| = \sqrt{x_1 ^2 x_2 ^2 + x_1^2 y_2^2 + x_2^2 y_1^2 +y_1^2 y_2^2}##

Which can easily be shown to be equal to ##|z_1||z_2|##.


So now that I have shown that ##|z_1 z_2| = |z_1| |z_2|## I can assume that the given statement holds true for a specific value of n such that n = k and I can move from step (2) to step (3) because I have proven the statement for k =2, correct?
 
  • #5
Tsunoyukami said:
Okay...so using ##z_1 = x_1 + iy_1## and ##z_2 = x_2 + iy_2##:

##|z_1 z_2| = |(x_1 + iy_1)(x_2 +iy_2)| = |(x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)|##

Now using definition of ##|z|## is such that ##|z| = |x + iy| = \sqrt{x^2 + y^2}## we find:

##|z_1 z_2| = |(x_1 x_2 - y_1 y_2) + i(x_1 y_2 + x_2 y_1)| = \sqrt{(x_1 x_2 - y_1 y_2)^2 + (x_1 y_2 + x_2 y_1)^2}##

Which, when simplified, yields:

##|z_1 z_2| = \sqrt{x_1 ^2 x_2 ^2 + x_1^2 y_2^2 + x_2^2 y_1^2 +y_1^2 y_2^2}##

Which can easily be shown to be equal to ##|z_1||z_2|##.


So now that I have shown that ##|z_1 z_2| = |z_1| |z_2|## I can assume that the given statement holds true for a specific value of n such that n = k and I can move from step (2) to step (3) because I have proven the statement for k =2, correct?

Yes.
 

1. What is proof by induction?

Proof by induction is a mathematical technique used to prove that a statement is true for all natural numbers. It involves proving the statement for a base case, typically n=1, and then showing that if the statement is true for any given value of n, it must also be true for the next value, n+1.

2. How is proof by induction used in complex variables?

In complex variables, proof by induction is often used to prove theorems and solve problems related to complex numbers and functions. It can be used to prove identities, inequalities, and other properties of complex numbers and functions.

3. What is the general process for using proof by induction?

The general process for using proof by induction involves three steps: 1) Proving the statement for the base case, typically n=1, 2) Assuming the statement is true for some value of n, and using this assumption to prove that it is also true for n+1, and 3) Concluding that the statement is true for all natural numbers based on the previous steps.

4. Can proof by induction be used for all types of mathematical statements?

No, proof by induction can only be used for statements that are true for all natural numbers. It cannot be used for statements that are only true for certain values or subsets of natural numbers.

5. How can I improve my skills in using proof by induction?

The best way to improve your skills in using proof by induction is to practice solving a variety of problems and theorems using this technique. You can also study and analyze example proofs to better understand the process and strategies used.

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