# All x for which 2^x < 8

1. Feb 4, 2012

1. The problem statement, all variables and given/known data

Find all x for which 2x < 8 (where x is a natural number).

2. Relevant equations

Those pesky 12 properties.

3. The attempt at a solution

This could be done by simply testing values of x, but that approach is not very scalable. What is the correct way to start this (using only the basic properties of numbers)? What is Spivak looking for? Thanks.

2. Feb 4, 2012

### A. Bahat

It's easy to find all x for which this holds by trial-and-error. Then prove that any other (larger) values of x give > instead of <.

3. Feb 4, 2012

As I said, I was wondering if there was another way. Trial and error doesn't seem very 'proofy' to me.

4. Feb 4, 2012

### Dick

I don't know the pesky 12 properties. But try writing it as 2^x<2^3. Any pesky properties to help with that?

5. Feb 4, 2012

12 Properties. But really, any number that makes you happy. I'm not really sure what to do with this. There is no basic property that says that i can bring down the exponents. This problem doesn't really matter in the long run, I was just curious as to what Spivak expects me to do with it.

6. Feb 4, 2012

### DivisionByZro

I think Dick's hint is spot on. I think his approach is what Spivak expected. Can you see what to do next?

7. Feb 4, 2012

### Jorriss

Dick's hint seems good, but if you can show by exhaustion that some x > some other number doesn't work, well, exhaustion is infact a valid proof. See, four color theorem.

8. Feb 4, 2012

### Office_Shredder

Staff Emeritus
Prove that the function 2x is an increasing function on the natural numbers directly: if f(x)=2x, show that f(x)<f(x+1) for all x. Then it's clear that 2x<23 if and only if x<3

9. Feb 4, 2012

Hello folks

@Office_Shredder: Is that induction? That is next chapter, so I wanted to hold off on that.

@Joriss: So when there is a contradiction, exhaustion is acceptable proof, but if there is no contradiction, hen it is not. Is that a true statement>

@DivisionByZro: Unfortunately, no. the only thing I can think to do with that is to write out what 2^x and 2^3 means:

2x < 23

2*2 x-number-of-times < 2*2 3-times

is that what you mean?

10. Feb 4, 2012

### Jorriss

No, exhaustion works for contradictions or when there are finite number of possibilities. I didn't say exhaustion is a good route here, just that when applicable it is a legitimate proof (which, if you already knew, my bad .