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Prove that the set $\{(x,y)\in\mathbb{R}^2:y=2x-1\}$ is a line in $\mathbb{R}^2$, which the points $a=(1,1) \text{ and } b=(0,-1)$ lies on

So as for the points, all the need to be done is to substitute the values and we get:

$a: 1=2*1-1\iff 1=1$ so $a$ lies on the line

$b: -1=2*0-1\iff -1=-1$ so $b$ lies on the line

But what do I need to show to prove that it is a line in $\mathbb{R}^2$?

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    What is your definition of "a line in $\mathbb R^2$"? – 2017-01-15
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    @MyGlasses There are many curves in the plane that are convex and aren't even close to a line, e.g. the parabola $\;f(x)=x^2\;$ . – 2017-01-15
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    @DonAntonio, I suppose they mean that the set of points which make up the line is a convex set. – 2017-01-15
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    @AntonioVargas I'm not sure I understand, but I think my prior comment remains the same... – 2017-01-15
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    @MyGlasses Topologically, without surprising, weird conditions, $\;C,\,\Bbb R^2\;$ are toplogically *the same*, and then one set is convex in one iff it is convex in the other one... – 2017-01-15
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    @DonAntonio, The term "convex set" has a specific meaning: https://en.wikipedia.org/wiki/Convex_set – 2017-01-15
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    @AntonioVargas Ok, now I see what you meant before. Thanks. Still, a disk (open or closed) is convex yet it is not a line... – 2017-01-15
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    @DonAntonio quite right indeed. – 2017-01-15
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    Can it be that I need to use the parametric representation of a line? Namely $r=ta+(1-t)b$? – 2017-01-15
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    @MyGlasses Neiter as a subset of $\;\Bbb R^2\;$ with the meaning Antonio meant... – 2017-01-15
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    @gbox Did you read and understand my answer below? – 2017-01-15
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    @DonAntonio using the fact there is one on line between two dots, we should look at the slope between the two dots and see if it is the same as the one in the function? – 2017-01-15
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    Another possibility: the graph of a one-variable *real* function is a line iff when taking **any** point $\;P\;$ , then for any *other* two points $\;A,\,B\;$ on the line we have that the angle between $\;\vec{PA},\,\vec{PB}\;$ is either $\;0\;$ or $\;\pi\;$ radians... – 2017-01-15

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I wonder how you're going to prove this, but one possible way is to show the slope between any two points on this curve is a constant number. Thus, if $\;(a,\,2a-1)\;,\;\;(b,\,2b-1)\;$ are two different points on the curve, the slope between them is

$$\frac{2b-1-(2a-1)}{b-a}=\frac{2(b-a)}{b-a}=2$$

If you want some more analytic-geometry-wise way, you can also take

$$\left\{\,(x,\,2x-1)\in\Bbb R^2\,\right\}=(0,-1)+t(1,2)\;,\;\;t\in\Bbb R$$

The right one above is the parametric equation of straight line in the plane, as it only has/needs one single direction vector

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    Is your definition of a line in $\mathbb{R}^2$ is having constant slop? Is it acceptable? – 2017-01-15
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    @MyGlasses I think so: it is the (infinite, of course) set of all points in the plane such that any two different ones have the same, constant slope. Vertical lines of the form $\;x=k=\;$ a constant need a slightly different definition, though. – 2017-01-15
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    $[x]$ is a line according above definition – 2017-01-15
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    @MyGlasses No, it's not, as if you take the points $\;(1.5,\,1),\,(1.6,\,1),\,(2,2)\;$ , then the slope between the first two is zero, yet the slope between the third one and the first one is *not zero* – 2017-01-15
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    well. what about $y=x$ with $x\in\mathbb{Q}$.? – 2017-01-15
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    @MyGlasses That does **not** cover all the point in $\;\Bbb R^2\;$ ...in your one before last comment. – 2017-01-15
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    What you defined in your last line is *only* a **constant** function = line with slope equal to zero. – 2017-01-15