I guess my answer is correct

[dollyayesha2345]

Give an example of each type of number.
a real number between [math]\frac59[/math] and [math]\frac69[/math]
My approach: Two-step technique
Step 1: As one real number is asked increasing that number with 1
[math]1+1=2[/math]
Step 2:Multiplying both the given fractions with[math]\frac22[/math][math]\frac59\times\frac22=\frac1018[/math](=10 upon18)
[math]\frac69\times\frac22=\frac1218[/math](=12upon18)
So one real number in between is 11/18

Is the answer correct?

 

[lev888]

Give an example of each type of number.
a real number between [math]\frac59[/math] and [math]\frac69[/math]
My approach: Two-step technique
Step 1: As one real number is asked increasing that number with 1
[math]1+1=2[/math]
Step 2:Multiplying both the given fractions with[math]\frac22[/math][math]\frac59\times\frac22=\frac1018[/math](=10 upon18)
[math]\frac69\times\frac22=\frac1218[/math](=12upon18)
So one real number in between is 11 upon 18

Is the answer correct?

Sorry, don't understand your reasoning and notation. And I don't see the answer.

 

[dollyayesha2345]

Sorry, don't understand your reasoning and notation. And I don't see the answer.

the Latex for fractions doesn't work properly my final answer is 11/18.

 

[lev888]

the Latex for fractions doesn't work properly my final answer is 11/18.

11/18 works. No idea what 1+1, etc is about. Is this how you find the average of 2 numbers?

 

[dollyayesha2345]

11/18 works. No idea what 1+1, etc is about. Is this how you find the average of 2 numbers?

in the question "to find is 1 real number" so the Step 1 of the two-step technique is to mention x+1=y (x the number required as per the question, y the number we get after the addition) So in my case 1(x)+1=2(y)

 

[lev888]

in the question "to find is 1 real number" so the Step 1 of the two-step technique is to mention x+1=y (x the number required as per the question, y the number we get after the addition) So in my case 1(x)+1=2(y)

Don't understand this explanation.
Here's what I would do:
Given numbers a and b, a<b, find a real number between them.
The average of a and b is exactly in the middle of the [a,b] interval. It equals (a+b)/2.
In your case: 5/9 + 6/9 = 11/9. And (11/9)/2 = 11/18.

 

[dollyayesha2345]

Don't understand this explanation.
Here's what I would do:
Given numbers a and b, a<b, find a real number between them.
The average of a and b is exactly in the middle of the [a,b] interval. It equals (a+b)/2.
In your case: 5/9 + 6/9 = 11/9. And (11/9)/2 = 11/18.

what is the name of this method that you have used?

 

[Harry_the_cat]

This is what I'd do:

\(\displaystyle \frac{5}{9} =\frac{10}{18}\)

\(\displaystyle \frac{6}{9} =\frac{12}{18}\)

Clearly, \(\displaystyle \frac{11}{18}\) lies between \(\displaystyle \frac{10}{18}\) and \(\displaystyle \frac{12}{18}\).

So, \(\displaystyle \frac{11}{18}\) lies between \(\displaystyle \frac{5}{9}\) and \(\displaystyle \frac{6}{9}\).

 

[dollyayesha2345]

This is what I'd do:

\(\displaystyle \frac{5}{9} =\frac{10}{18}\)

\(\displaystyle \frac{6}{9} =\frac{12}{18}\)

Clearly, \(\displaystyle \frac{11}{18}\) lies between \(\displaystyle \frac{10}{18}\) and \(\displaystyle \frac{12}{18}\).

So, \(\displaystyle \frac{11}{18}\) lies between \(\displaystyle \frac{5}{9}\) and \(\displaystyle \frac{6}{9}\).

and the name of your method is?

 

[lev888]

what is the name of this method that you have used?

I described exactly what I was doing. The method is called solving math problems.

 

[Deleted member 4993]

Give an example of each type of number.
a real number between [math]\frac59[/math] and [math]\frac69[/math]
My approach: Two-step technique
Step 1: As one real number is asked increasing that number with 1
[math]1+1=2[/math]
Step 2:Multiplying both the given fractions with[math]\frac22[/math][math]\frac59\times\frac22=\frac1018[/math](=10 upon18)
[math]\frac69\times\frac22=\frac1218[/math](=12upon18)
So one real number in between is 11/18

Is the answer correct?

Please follow the method that you had followed in:

I got no clue

Question: Find a real number between 1 and\sqrt3 My approach(which is wrong): \frac{a+b}{2} \frac{1+1.732}{2} =\frac{2.732}{2} =1.366 So a real number between 1 and\sqrt3 is 1.3. Please tell me whether my approach and answer is correct or not. If not then do guide me on how to solve this.

www.freemathhelp.com

I do not know the official name of that method - but I call it "finding the middle by average".

 

[Otis]

\frac59\times\frac22=\frac1018

the Latex for fractions doesn't work properly

Hi Dolly. You didn't type the delimiters. The syntax for the \frac{}{} command is to place the numerator and denominator each inside a set of curly braces, like this:

\frac{10}{18}

PS: There's a LaTeX thread on the News board with examples and tutorial links.

?

 

[Harry_the_cat]

and the name of your method is?

I don't think it has a name. I'd call it "converting to a larger denominator to make room between the numerators"!