Question

if sqrt2 lies between (x+3)/x and (x+4)/(x+1) , find the integral values of x.

Answer 1

integral values mean positive integral?
or both positive integral and negative integral?

Answer 2

integral= integer so both + and -

Answer 3

it's an interesting question\[\frac{x+3}{x}< \sqrt{2} < \frac{x+4}{x+1}\]\[1+\frac{3}{x} < \sqrt{2} < 1+\frac{3}{x+1}\]

Answer 4

x can not be a great integer

Answer 5

so how to do it?

Answer 6

Play with some values to see what does or doesn't fit.

For the left hand of your inequality, you know it has to be an integer that makes

1 + 3/x less than sqrt 2 (~1.41)

so x would have to be at least 8 in the positive direction (1 + 3/7 > sqrt 2);

anything in the negative direction makes the inequality true true, since anything subtracted from 1 will be less than sqrt 2.

For the right hand side, if we plug in 8 - the value we know makes the left hand side correct - it gives us (1 + 3/9) ≈ 1.333

1.33 < 1.41, so 8 cannot be an integer value of x, nor can anything greater than 8, since it only makes the term smaller. That means there are no positive integer solutions that satisfy the inequality.

Next you can look at the negative integers. x has to be greater than -1, otherwise you're dividing by zero. Any other value of x in the negative direction, however, also does not satisfy the inequality, again because any value that's subtracted from 1 will be less than 1, therefore less than sqrt 2

. ie, x = -4
1 + 3/(-4 + 1) = 1-1 = 0 < sqrt 2

Lastly, cannot be zero because of the left hand side.

Therefore, since there exist no positive or negative integer solutions, just like mukushla said, x cannot be a great integer. :)

Answer 7

"Next you can look at the negative integers. x has to be greater than -1,"

~ ~ has to be less than -1