Question

hi help in basic arithmetic algebra

Answer 1

lol nice intro ;)

Answer 2

:).... can u help @lgbasallote

Answer 3

You need to post your question in order for people to answer it.

Answer 4

i had posted the image @Wired ... see it

Answer 5

@mathslover

Answer 6

How far have you gotten on your own?

Answer 7

nice reply @lgbasallote .....this means that you can not help me ....

Answer 8

cual es la pregunta ¿

Answer 9

yeah.. im not knowledgeable in basic arithmetic algebra :(

Answer 10

The image just popped up.

Answer 11

i think this can be proved like this :

If x > 1 then we can say that 1.x > 1 ... So n = 1 and i think it is done

but i think this is not a proper way

Answer 12

@KingGeorge ...

Answer 13

That proves it if \(x\ge1\). We also need to consider the case where \(0<x<1\).

Answer 14

we can apply the archimedean property to number 1/x ...

Answer 15

By definition, there exists some \(x^{-1}>0\) (i.e \(\frac{1}{x}\)) such that \[x\cdot x^{-1}=1\]Now, take \(n\in\mathbb{N}\) such that \(n>x^{-1}\). This means that \[x\cdot n>x\cdot x^{-1}=1\]Thus, \(x\cdot n>1\).

Answer 16

ok thanks a lot...

Answer 17

You're welcome.