Proof without calculation

Question

Show that the product of two of the numbers $(65^{1000} - 8^{2001} + 3^{177}), (79^{1212} - 9^{2399} + 2^{2001})$ and $(24^{4493} - 5^{8192} + 7^{1777})$ is non-negative, without actually evaluating the numbers.

P.S. I have found by calculation that all the three numbers are positive, but that does not solve the problem of proving without calculation.

Thanks in advance.

Answer 1

I think the trick is that the question just asks for a proof that there exist two of the numbers such that their product is non-negative. In principle there could be other products that were negative.

Now, if at least two of the numbers are non-negative, then their product is non-negative too.

If less than two of them are non-negative, there must be at least two negative numbers among them ...

Answer 2

Consider for instance $65^{1000} - 8^{2001} + 3^{177}$. Note that $65>64=8^2$ and so $65^{1000} > 8^{2000}$, but it'd be better if we had $65^{1000} > 8^{2001}$ because then $65^{1000} - 8^{2001} + 3^{177}$ would definitely be positive. So we need a finer estimate for $65^{1000}$. Here is one: $65^{1000} = (64+1)^{1000} = 64^{1000}+1000\cdot 64^{999}+\cdots > 64^{1000}+7\cdot 64 \cdot 64^{999} = 8 \cdot 64^{1000} = 8^{2001}$.

The other two numbers are positive in the same fashion but you need a different argument.

Answer 3

Unless I did a mistake in the calculations, this should solve the last one.

$$24^{4493}=2^{3*4493}*3^{4493}$$

Now using

$$2^7 \geq 5^3 \,;\, 3^3 \geq 5^2 \,,$$

we get

$$2^{3*4493}*3^{4493} \geq (2^7)^{1925}(3^3)^{1497} \geq 5^{1925*3+1497*2}=5^{8769}$$

And here is the other

$$(\frac{79}{81})^{20}= (1-\frac{2}{81})^{20} \geq 1-\frac{40}{81} \,.$$

by Bernoulli

Thus

$$(\frac{79}{81})^{100}\geq \frac{1}{2^5} \geq \frac{1}{79} \,.$$

Thus

$$79^{101} \geq 81^{100} \,.$$

And hence

$$79^{1212} \geq 81^{1200} $$

The positivity of the second term is an immediate consequence of this....

P.S. Edit The last inequality also follows by this idea:

We show that

$$24^{4493} \geq 25^{4096}$$

$$(\frac{24}{25})^{12} \geq 1-\frac{12}{25} \geq \frac{1}{2}$$

$$(\frac{24}{25})^{12*342}\geq \frac{1}{2^{342}} \geq{1}{24^{389}} \,.$$

Answer 4

Building on @henning-makholm this is a question in the 8th Ed textbook by Rosen. I solved it as follows:

Let $P(x,y)$ represent the predicate $xy > 0$, and the domain be $a,b,c$ represented by $(65^{1000}−8^{2001}+3^{177}),(79^{1212}−9^{2399}+2^{2001}),(24^{4493}−5^{8192}+7^{1777})$ respectively.

• We can conclude that $abc \ne 0$
• Case i) At least 2 of $a,b,c$ are non-negative
• Case ii) At least 2 of $a,b,c$ are negative

We know that the sum of two integers with the same sign will result in a non-negative number.

$\therefore \forall x \exists yP(x,y)$ such that $xy>0$.

This is a nonconstructive proof.

Answer 5

We are being asked to show that the product of two numbers out of three is non-negative. A number $n$ is non-negative if $n=0$ or $n>0$. Consider, then, two cases:

1. At least one of the numbers is zero. If this is true, then multiplying it by any of the other numbers will yield zero--which is non-negative.
2. None of the numbers are zero. This means each number is either positive or negative. A product will be non-negative when either both factors are positive or both are negative. If one of the numbers is positive, then either (a) one of the remaining two is positive or (b) both of the remaining two are negative; in either case, we can pick two numbers whose product will be non-negative. We can apply similar reasoning if one of the numbers is negative.

This is a non-constructive proof.

Remarks:

Case 2 can be visualized with something akin to a truth table. If we have three numbers $a, b, c$ that are either positive or negative, then we get $8$ possibilities:

a	b	c
+	+	+
+	+	-
+	-	+
+	-	-
-	+	+
-	+	-
-	-	+
-	-	-

As we can see from the table, in every case we can find either two positive or two negative numbers to multiply.