I guess my answer is correct

dollyayesha2345

Junior Member
Joined
Oct 28, 2021
Messages
57
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?
 
Last edited:

lev888

Elite Member
Joined
Jan 16, 2018
Messages
2,677
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

Junior Member
Joined
Oct 28, 2021
Messages
57
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

Elite Member
Joined
Jan 16, 2018
Messages
2,677
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

Junior Member
Joined
Oct 28, 2021
Messages
57
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

Elite Member
Joined
Mar 16, 2016
Messages
3,135
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

Junior Member
Joined
Oct 28, 2021
Messages
57
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?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
25,895
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 do not know the official name of that method - but I call it "finding the middle by average".
 
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