Proofs Question (1 Viewer)

CM_Tutor

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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

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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 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 .
 

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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:








as required
 

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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.







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
 

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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.
 
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