pine-apple01320
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Hi, I was wondering how to prove the following question?
10^9 < 9^10
10^9 < 9^10
-
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Proofs Question (1 Viewer)
- Thread starter pine-apple01320
- Start date
Theorem: 
This sort of problem can be approached by calculus (as in the link provided by Drdusk), but calculus need not be used.
In this case, "proof" by calculator makes the result that obviously true as 
is one billion and 
is over three billion. However, such an answer is not helpful in general as this type of response can be easily avoided by making the numbers larger. One alternative (and usable for larger numbers) is to re-write with a common base.
Proof 1 (common base):
We know from index laws that
leads to the conclusion that 
provided that the base a is real and 
. So, if we can re-write 
in the form 
, the result is proved if 
(or similarly, re-write 
as 
and show that 
).
Or
---
This proof is similar to a proof by caluclator, but it would allow you to prove that
as 
and 
. The approach won't generalise to pronumerals, though, so looking at (say) 
and ^a)
for an appropriate domain of a is not approachable in this way. (I note a suitable domain of a as is larger for 
, the statements are equal for 
, and is smaller for 
.
This sort of problem can be approached by calculus (as in the link provided by Drdusk), but calculus need not be used.
In this case, "proof" by calculator makes the result that
Proof 1 (common base):
We know from index laws that
Or
---
This proof is similar to a proof by caluclator, but it would allow you to prove that
These inequalities can make use of the techniques in the "Proof" part of the syllabus and using the binomial theorem.
Proof 2 (using the binomial theorem):
Proving that
is negative is equivalent to proving that 
. Here, we have:
^{10}})
)
)
)
)
as required
Proof 2 (using the binomial theorem):
Proving that
Proof 3 (using the binomnial theorem):
The binomial theorem can also be used to examine the ratio of the two sides, such as by proving that
. This leads to 
provided that we know the RHS to be positive.
^9 \times \frac{1}{9} = \bigg(1 + \frac{1}{9}\bigg)^9 \times \frac{1}{9})
 \times \frac{1}{9})
 \times \frac{1}{9})
 \times \frac{1}{9})
Approximating, we know that every term in this sum is positive and all of the last six terms)
is much less than 1, and so we can restrict the possible values of the ratio to a narrow range:
It is clear that
and both of these terms is positive, hence as required
The binomial theorem can also be used to examine the ratio of the two sides, such as by proving that
Approximating, we know that every term in this sum is positive and all of the last six terms
It is clear that
The methods of proof by contradiction can also be used:
Proof 4 (by contradiction):
Assume that the result is false, and thus that
Since the function
strictly increases throughout its domain (for any positive base) and is positive for 
, we can take logs of both sides here (where both terms are clearly greater than 1) without altering the validity of the inequality:
From the calculator, this is
which is clearly false and since every statement from the assumption follows logically and yet leads to a contradiction, the only valid conclusion is that the assumption must be false. It follows immediately that the required result is true.
Proof 4 (by contradiction):
Assume that the result is false, and thus that
Since the function
From the calculator, this is
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