Dealing with the inquiries of attributes in eleventh course my tennis2007.orgs teacher claims that square source of a real number is constantly positive. Just how is that possible?


*

*

Given a positive real number $a$, there are two solutions to the equation $x^2 = a$, one is positive, and also the other is negative. We denote the positive root (which we often call the square root) by $sqrta$. The an unfavorable solution that $x^2 = a$ is $-sqrta$ (we know that if $x$ satisfies $x^2 = a$, then $(-x)^2 = x^2 = a$, therefore, due to the fact that $sqrta$ is a solution, therefore is $-sqrta$). So, because that $a > 0$, $sqrta > 0$, but there are two services to the equation $x^2 = a$, one positive ($sqrta$) and also one negative ($-sqrta$). Because that $a = 0$, the two options coincide v $sqrt0 = 0$.

You are watching: Positive numbers have negative square roots true or false


share
point out
follow
answered may 26 "14 in ~ 1:01
*

Michael AlbaneseMichael Albanese
85.4k1717 yellow badges160160 silver- badges405405 bronze title
$endgroup$
6
| show 1 much more comment
7
$egingroup$
It is simply a notational matter. By convention, for hopeful $x$ (real clearly), $sqrtx$ denotes the positive square source of the genuine number $x$. Similarly we agree by way of notational convention the $-sqrtx$ is the an adverse square source of $x$. That course, every optimistic real number, $x$, has actually two square roots, $sqrtx$ and $-sqrtx$, confident and negative real numbers respectively.

I problem sometimes about what it s okay taught by means tennis2007.orgematics in an additional school this days.


share
cite
follow
edited may 26 "14 in ~ 7:02
*

usertennis2007.org
3,3811818 silver badges3333 bronze title
answered might 26 "14 at 6:34
*

Richard GayleRichard Gayle
7111 bronze argorial
$endgroup$
1
add a comment |
2
$egingroup$
Technically this declare is wrong. He could say, "The square root of a optimistic number is confident (by definition)". E.g. Because that 0 you get $sqrt0=0$ which is neither confident nor negative. And for negative numbers you also get complicated solutions which are neither confident nor an adverse nor 0.

The identify article and also the singular in "the square root" is also important to imply the conventional meaning of $sqrt$. But much more correctly he need to say "the major square root", due to the fact that tennis2007.orgematically the expression "the square root" doesn"t make sense, since there room two different roots in general.


share
mention
follow
answered may 26 "14 at 18:18
David OngaroDavid Ongaro
23022 silver badges66 bronze title
$endgroup$
include a comment |
2
$egingroup$
I believe your man is comes from assuming that if $a^2 = b$ then $sqrtb =a$, yet this is actually not the case. The correct form of this would be $sqrtb = vert a vert$. Because of this, $a^2 = b = (-a)^2$ however $sqrtb = vert a vert eq -vert a vert$. This can be proven by contradiction:

If us were come say that $a=sqrtb=-a$, the would therefore imply that $a = -a$, and for example $1= sqrt1 = -1 implies 1=-1$ i m sorry we recognize to be false. This contradiction walk not present up once saying $a^2 = b = (-a)^2 implies a^2 = (-a)^2$, because, if us were to usage the very same example, we would gain $1^2 = 1 = (-1)^2 implies 1^2 = (-1)^2 implies 1 = 1$, which is true (since by definition, any kind of number squared must it is in positive). This is why, if you to be to advice $a^2 = b$, girlfriend would gain two possible systems for $a$ (one positive, and also one negative). However, if you to be to evaluate the equation $a = sqrtb$, $a$ deserve to only have actually one equipment at any kind of given time, and also for convention, a square source was identified to constantly be positive. This is crucial distinction since it enables us come look in ~ the equation $4=a^2$, and also find the $a=2oplus a=-2$, thus staying clear of the contradiction $a=2 wedge a=-2 implies 2=-2$ by speak $2^2=4=2^2 implies 2 = 2 oplus (-2)^2=4=(-2)^2 implies -2=-2$. This idea could seem to gain lost once graphing equations such together a circle. The equation $x^2 + y^2 = 1$ appears to have actually 2 $y$ worths for every $x$, and 2 $x$ values for every $y$. This can be far better understood by instead looking at the parametric equation for a circle: $x=cos(t); y=sin(t)$. For any kind of given worth of $t$, over there is just one equivalent value the $x$, and only one matching value the $y$. If you are offered a value for $x$, and told to deal with for $t$, the most you deserve to do is find possibilities the $t$, since the $(x,y)$ clues on the graph repeat themselves every $2pi*t$. The very same idea is true for square roots. Once you square a number, it constantly creates a positive number, because of this it is difficult to reverse definitively. The most we deserve to do is say the there room two possibilities the what the initial number was. Because that convention, it has actually been established that for an equation $a^2 = b$, wherein $sqrtb=c$, us say the $c=vert a vert$. That would work just too to defined a square source by $c=-vert a vert$, however I guess: v the tennis2007.orgematicians that decided on it liked working with confident numbers more.

The suggest in all of this was to simply establish that taking the square source of a squared number, does no reverse its exponent, due to the fact that it can not be reversed definitively. As user86418 put it:

If a and b are genuine numbers, then the problems $a^2=b$ and also $a=sqrtb$ are not logically equivalent; the 2nd implies the first, yet not conversely.

See more: How Far Is Clearwater Beach From Fort Myers Beach, Fl, How Far Is Clearwater From Fort Myers Beach

Therefor, because that the objectives of convention, a square root has been characterized to it is in the absolute worth of the original number the was squared. This why, if girlfriend plug the features $y^2 = x$ and also $y = sqrtx$ into a graphing calculator or Wolfram Alpha, you will uncover that you get two very different feather graphs. Notification how the the graph the $y=sqrtx$ never ever goes below the $x$-axis. Had actually a square source been identified as constantly negative, the graph that $y=sqrtx$ would merely be flipped around the $x$-axis.