I need help trying to figure out how I can prove a statement like this. So looking at this I can conclude that this statement is of the form $P \Rightarrow Q$

$$ P: x^4 + 5x + 1 < 27 $$ $$ Q: x < 2 $$

I wanted to try and prove this by contrapositive , so this state would become

If $X \geq 2$ then $x^4 + 5x + 1 \geq 27$

Over here I was not sure if I could plug in the value 2 or anythin greater than 2 to see if this is true. plugging in 2 gets me $2^4 + 5 \times 2 + 1 = 27$ and since $27 \geq 27$ this statement is True.

Am I allowed to prove it like this? Is there a different way to prove a question like this?

  • No, you have to prove for a general $x\geq 2$. It might work for a specific value but not in general. That is the problem with your approach – Heisenberg Oct 20 at 22:45
  • If $x \ge 2$ then $x^4 + 5x +1 \ge 16+10+1 = 27$. – copper.hat Oct 20 at 22:47
  • You can use that for $c\ge0$, $a\ge b$ implies $ac\ge bc$. So for $x\ge 2$ (and thereby $x>0$) we have $x^2\ge 2x\ge 4$, $x^3\ge 2x^2\ge 8$, $x^4\ge 2x^3\ge 16$, and hence by addimg the inequality $x^4+4x+1\ge 16+10+1=27$ indeed – Hagen von Eitzen Oct 20 at 22:48
up vote 3 down vote accepted

Suppose $x\geq2$. Then $x^4\geq16\implies x^4 +5x\geq16+5x\geq 16+...$. Got the hint?

  • Are you trying to make the left side look like what we need to prove? – Rufyi Oct 20 at 23:01
  • If I start from x^4 >= 16, I would get x^4 + 5x + 1 >= 16 + 5x + 1, can i plug in 2 in for 5x on the right side? – Rufyi Oct 20 at 23:05
  • Yes exactly!... – Heisenberg Oct 20 at 23:05
  • Can I do what I said in the second comment? – Rufyi Oct 20 at 23:09
  • Yes when you plug in 2, you get 27 on the right side thus proving $x^4+5x+1\geq 27$ – Heisenberg Oct 20 at 23:11

A marginally different take:

Let $p(x) = x^4+5x-26$, we would like to show that if $p(x)<0$ then $x <2$.

Note that $p(2) = 0$ and so synthetic division gives $p(x) = (x-2)(x^3+2x^2+4x+13)$.

In particular, note that $x^3+2x^2+4x+13$ has no positive roots. Hence if $x \ge 0$, we see that $p(x)$ and $x-2$ have the same sign, hence if $p(x) < 0$ we must have $x < 2$.

Trying to show the contrapositive is a valid method, but you aren't allowed to just give an example. For example, if you wanted to prove that

$$x\geq 2\implies x^4+5x+1\leq 27,$$

which is false for example at $x=0$, you can't just give the example of $x=2$.

One hint for trying to prove this is to show that $f(x)=x^4+5x+1$ is increasing over the positive real numbers - in other words, assume that $0<a<b$ and show that $f(b)>f(a)$.

If $x\ge 2$ is it true that $x^4 \ge 16$?

If $x \ge 2$ is it true that $5x \ge 10$?

So if $x \ge 2$ is it true that $x^4 + 5x + 1 \ge 16 + 10 + 1 = 27$?

...

It could get tedious to go to axioms and prove that if $x\ge 2 > 0$ then $x^4 \ge 2x^3 \ge 4x^2 \ge 8x \ge 16$ via the axiom: if $a > 0$ and $m < n$ then $am < an$ (applied three times); and $5 > 0$ so $x\ge 2 \implies 5x \ge 2*5 =10$ by the same axiom; and by the axiom $a + c > b + c \iff a > b$ then $x^4 + 5x + 1 \ge x^4 + 10 + 1 \ge 16+ 10 + 1=27$. But I think it is safe to assume if we had to prove those we could.


This may be tedious over kill but:

If $x \ge 2$ then let $d = x - 2 \ge 0$. Then $x = 2 + d$

Now because $d \ge 0$ then $d^k \ge 0$.

So $x^4 = (2 + d)^4 = 2^4 + 4*2^3*d + 6*2^2*d^2 + 4*2*d^3 + d^4 \ge 2^4$

And $5x = 5*(2+d) = 5*2 + 5*d \ge 5*2$

So $x^4 + 5x + 1\ge 2^4 + 5*2 + 1 = 27$.

That should do it.....

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