Is this induction proof valid?

  • Thread starter 1MileCrash
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  • #1
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Homework Statement



My induction proof was really easy and simple, and people were having problems and doing substitutions and other complicated things. Normally I'd take pride in making a simpler proof but, there are really brilliant people in this class and I'm just starting to get the grasp of proofs, so it's not comforting.

Prove that for all real numbers:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Homework Equations





The Attempt at a Solution



Let S be the set of natural numbers such that:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Let n = 1

[itex](1+\frac{1}{2})^{1} = 1.5[/itex]

[itex]1 + \frac{1}{2} = 1.5[/itex]

Therefore, 1 E S

Assume this is true for n = k

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Check if this implies truth for n = k + 1
Left side:
[itex](1+\frac{1}{2})^{k+1}[/itex]

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2})[/itex]

Right side:

[itex]1 + \frac{k+1}{2}[/itex]

[itex]1 + \frac{k}{2} + \frac{1}{2}[/itex]


[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Is a true statement because our initial assumption asserts that

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Here, the greater term is multiplied by 1.5, and the lower term has .5 added to it. A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal. Since the first term is greater than the second, and is operated on a way such that a greater result is yielded even if the two were equal,

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Must be true.


Therefore S is inductive.

Therefore S = Natural Numbers.
 

Answers and Replies

  • #2
34,825
6,568

Homework Statement



My induction proof was really easy and simple, and people were having problems and doing substitutions and other complicated things. Normally I'd take pride in making a simpler proof but, there are really brilliant people in this class and I'm just starting to get the grasp of proofs, so it's not comforting.

Prove that for all real numbers:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Homework Equations





The Attempt at a Solution



Let S be the set of natural numbers such that:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Let n = 1

[itex](1+\frac{1}{2})^{1} = 1.5[/itex]

[itex]1 + \frac{1}{2} = 1.5[/itex]

Therefore, 1 E S

Assume this is true for n = k

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Check if this implies truth for n = k + 1
Left side:
[itex](1+\frac{1}{2})^{k+1}[/itex]

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2})[/itex]

Right side:

[itex]1 + \frac{k+1}{2}[/itex]

[itex]1 + \frac{k}{2} + \frac{1}{2}[/itex]


[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]
I think you are hand-waving here. Start with the left side and show that it is >= 1 + (k + 1)/2.
Is a true statement because our initial assumption asserts that

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]


Here, the greater term is multiplied by 1.5, and the lower term has .5 added to it. A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal. Since the first term is greater than the second, and is operated on a way such that a greater result is yielded even if the two were equal,

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Must be true.


Therefore S is inductive.

Therefore S = Natural Numbers.
 
  • #3
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement



My induction proof was really easy and simple, and people were having problems and doing substitutions and other complicated things. Normally I'd take pride in making a simpler proof but, there are really brilliant people in this class and I'm just starting to get the grasp of proofs, so it's not comforting.

Prove that for all real numbers:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Homework Equations





The Attempt at a Solution



Let S be the set of natural numbers such that:

[itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

Let n = 1

[itex](1+\frac{1}{2})^{1} = 1.5[/itex]

[itex]1 + \frac{1}{2} = 1.5[/itex]

Therefore, 1 E S

Assume this is true for n = k

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Check if this implies truth for n = k + 1
Left side:
[itex](1+\frac{1}{2})^{k+1}[/itex]

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2})[/itex]

Right side:

[itex]1 + \frac{k+1}{2}[/itex]

[itex]1 + \frac{k}{2} + \frac{1}{2}[/itex]


[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Is a true statement because our initial assumption asserts that

[itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

Here, the greater term is multiplied by 1.5, and the lower term has .5 added to it. A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal. Since the first term is greater than the second, and is operated on a way such that a greater result is yielded even if the two were equal,

[itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

Must be true.


Therefore S is inductive.

Therefore S = Natural Numbers.

The flaw is this. "A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal." You don't have the "same number" on both sides. Just write (1+1/2)^k*(1+1/2) as (1+1/2)^k+(1+1/2)^k/2 and try that again.
 
  • #4
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Why does it matter if they aren't the same number? The left is greater than the right by induction hypothesis.
 
  • #5
Dick
Science Advisor
Homework Helper
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Why does it matter if they aren't the same number? The left is greater than the right by induction hypothesis.

The problem is in your exposition. The language is confusing. I think you should go back to taking pride in making a simpler proof. It's not that hard. You want to show (1+1/2)^k*(1+1/2)=(1+1/2)^k+(1+1/2)^k/2>=1+k/2+1/2.
 
  • #6
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What I am saying is basically, if I assume that x is greater than y, then x + 2 is greater than y + 1.

Of course, +1 and +2 are examples of operations.
 
  • #7
Dick
Science Advisor
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What I am saying is basically, if I assume that x is greater than y, then x + 2 is greater than y + 1.

Of course, +1 and +2 are examples of operations.

Sure, now why don't you try to clear up the original proof?
 
  • #8
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40
Ok, I'll take a shot. So you're telling me that I should translate the jist of it into math?
 
  • #9
Dick
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Ok, I'll take a shot. So you're telling me that I should translate the jist of it into math?

Yes, show (1+1/2)^k*(1+1/2)=(1+1/2)^k+(1+1/2)^k/2>=1+k/2+1/2. I've been repeating that for a while. It's pretty direct.
 

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