Usually the square root of a number is much smaller than the number you start with?

Question

Usually the square root of a number is much smaller than the number you start with, because you can multiply the two square roots together and come up with you original number. BUT, brain quiz: Why is it that, when you take the square root of a decimal, like .7 you come up with LARGER numbers than the square root? Like, the square root of .7 is .84. It is very strange that a square root is larger than the original number.

Answer 1

The thing is, you are usually taking the sqrt of whole numbers, not decimals.

But then again, decimals can be turned into fractions, which you can take the sqrt of, and it still follows that the sqrt of a number is smaller than what you started with.



As with the example you did, .7 = 7/10

sqrt(7/10) = {sqrt(7)} / {sqrt(10)}

sqrt(7) = 2.65

sqrt(10) = 3.16

sqrt(7/10) = sqrt(.7) = 2.65/3.16 = .84



When you break it down, the sqrt is still smaller than the original number.

Answer 2

The square root of a number is only less than that number if that number was more than 1.

3*3 = 9 so sqrt(9) = 3



If the original number is less than 1 its square root will always be larger.

(1/3)*(1/3) = 1/9 so sqrt(1/9) = 1/3



When you say "usually" you are just reflecting the fact that most of the numbers you have had to find roots of have been larger than one up until now. You have got used to the root-means-smaller idea, hence this fact, (that was always there), seems strange.



The square root of 1/10,00 is 1/100. You can make as many of these as you like.



Regards - Ian

Answer 3

is because the square root mustb be multiplied by the same value to get the square of a number. This means one is multiplying a fractio by a fraction, like 1/2 x 1/2 =1/4; In short, one is is taking a fraction of a fraction when squaring a fraction.and when one does this the result will always be a smaller number. Example 9/16 =3/4 x3/4=9/16 because one is taking 3/4 of 3/4=9/16

Good Luck. Hope this helps.

Answer 4

this methodology became taught until eventually now we had calculators: think your variety is 155236 (which isn't an obtrusive sq.) Separate the digits into pairs and set up a "branch" challenge: ?15 fifty two 36 be sure the biggest sq. smaller than the leftmost digit or pair, and write its sq. rooot above: . . 3 ?15 fifty two 36 sq. the variety and subtract, "bringing down" the subsequent pair: . . 3 ?15 fifty two 36 . . 9 . . 6 fifty two to locate the nexct digit, multiply the respond with the aid of 20 and use the product for a divisor: Use a _ rather of a nil, although because of the fact we can exchange the final digit: . . . 3 . ?15 fifty two 36 . . . 9 6_ |6 fifty two Write the quotient above, and interior the gap presented with the aid of the _. Then multiply with the aid of this digit . . . 3 . 9 . ?15 fifty two 36 . . . 9 6_ |6 fifty two . . . 3 . 9 . ?15 fifty two 36 . . . 9 sixty 9 |6 fifty two . 9 |6 21 Subtract, back "bringing down" the subsequent pair. Make a clean divisor with the aid of doubling the respond and including _: . . . 3 . 9 . ?15 fifty two 36 . . . 9 sixty 9 |6 fifty two . 9 |6 21 78_ | 31 36 (observe that the recent divisor is likewise the sum of the previous one and the multiplier.) Divide back and carry out the multiplication as until eventually now: . . . 3 . 9 . 4 . ?15 fifty two 36 . . . 9 sixty 9 |6 fifty two . 9 |6 21 784 | 31 36 . . . 4 | 31 36 Aince the variety became a sq. we've our answer. If the variety weren't a appropriate sq., the approach might nicely be persevered until eventually you attain the wanted decimal accuracy. For cube roots, there is often Newton's iterative approach: (Y - x^3)/(3x^2) = x1 - x x1 = (Y - x^3)/(3x^2) + x finding out on a smaller variety for an occasion, 10 all of us comprehend that 8 = 2^3, so attempt 2.a million for a customary guess x1 = (10 - 2.a million^3)/(3*2.a million^2) + 2.a million 2.a million^2 = 4.40-one 4.40-one*2.a million = 8.80 two + .441 = 9.261 3*4.40-one = 13.23 x1 = (10 - 9.261)/(13.23) + 2.a million x1 = (0.739)/(13.23) + 2.a million . . . . . . 0.0568 1323)seventy 3.ninety . . . . . sixty six.15 . . . . . . 7.750 . . . . . . 6.615 . . . . . . a million.1350 x1 = 2.a million + 0.0568 = 2.1568 . . . As you will see, this gets tedious exceedingly quickly. because of the fact that we've the respond bracketed, that's quicker & much less demanding to motel to the averaging approach from this factor.

Answer 5

Any number that is 0 < x < 1 the denominator will be larger than the numerator. Since the denominator is larger than the numerator, the denominator will grow faster when multiplied by another number and therefore results in a smaller number. This is the same reason why a number greater than 1 will result in a larger number. The numerator in a number greater than 1 will grow faster and result in a larger number.



If, 0 < x < 1, then sqrt( x ) > x

If, 1 < x < oo, then sqrt( x ) < x

If, -oo < x < 0, then sqrt( x ) is imaginary.



I hope this helps you. Have a great day.

Answer 6

The square root of a number is always closer to 1.

Answer 7

Not strange at all.

The graph of f(X) = √x is above the graph of g(x) = x in the interval (0,1). They intersect at 0 and 1. And f < g only for x > 1.



When you multiply by a number < 1, you get a smaller number, so of course you get what you described for x<1

Answer 8

Think of them as fractions ie .7 = 7/10

A good way to visulise this is to take the value 49/100

If you take the SqR you get 7/10, hence getting higher value than you start with