# Thread: Correct way to answer calculus question

1. ## Correct way to answer calculus question

Heres a question from a Foundations of Calculus past paper and i was wondering whether the answer given is really needed...

Q: Prove That Sup A = 10, where $A = (-\infty ,10)$

A: Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2 (Properties at end of question).
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.

Let $x = \frac{1}{2}(10 + \alpha)$.

Then $x < \frac{1}{2}(10 + 10) = 10$ so $x \in A$.
On the other hand $\alpha = \frac{1}{2}(\alpha + \alpha) < \frac{1}{2}(10 + \alpha) = x$.

Therefore, $\alpha$ is not an upper bound for A, which contradicts property 1. Absurd.

Therefore, $\alpha \geq 10$ so $\alpha = 10$.

Properties: A real number $\alpha$ is the supremum of a non-empty set of real numbers A if and only if...
1; $x \leq \alpha$ for all $x \in A$; and
2; if $x \leq \gamma$ for all $x \in A$, then $\alpha \leq \gamma$

And heres my thought... Cant you just use the Archimedian Property of Rational Numbers to say that...

Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2.
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.

By the A.P.O.R.N there exists an x s.t. $\alpha < x < \gamma$. Therefore, $\alpha$ is not an upper bound. Absurd... etc...

2. Originally Posted by Deadstar
... heres my thought... Cant you just use the Archimedian Property of Rational Numbers to say that...

Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2.
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.

By the A.P.O.R.N there exists an x s.t. $\alpha < x < \gamma$. Therefore, $\alpha$ is not an upper bound. Absurd... etc...
That is essentially the same argument as in the given solution. The only diference is that your method brings in the Archimedean axiom to prduce a rational number x, whereas the given solution simply defines x=½(10+α). Since there is no need for x to be rational, this seems a bit more straightforward than using the Archimedean axiom. But both methods are equally acceptable as proofs.