inequation problem: m, n>0, m, n integers, √(7)-m/n>0: prove that √(7)-m/n>1/mn

[akramzine]

inequation problem: m, n>0, m, n integers, √(7)-m/n>0: prove that √(7)-m/n>1/mn

please help me solve this problem:
m>0 and n>0 .m and n being integers
√(7)-m/n>0
prove that √(7)-m/n>1/mn

 

[Deleted member 4993]

please help me solve this problem:
m>0 and n>0 .m and n being integers
√(7)-m/n>0
prove that √(7)-m/n>1/mn

As you posted the problem:

m>0 and n>0 .m and n being integers

√(7) - (m/n) > 0

prove that

√(7) - (m/n) > (1/m) * n

Is this correct?

 

[akramzine]

prove that √(7) - (m/n) > (1/m*n)

 

[Deleted member 4993]

prove that √(7) - (m/n) > (1/m*n)

which is same as (posted before):

√(7) - (m/n) > (1/m) * n

or you mean,

√(7) - (m/n) > 1/(m * n)

 

[akramzine]

or you mean,

√(7) - (m/n) > 1/(m * n)

Yes this is what I mean

 

[Deleted member 4993]

or you mean,

√(7) - (m/n) > 1/(m * n)

Yes this is what I mean

That being settled:

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[akramzine]

I'm sorry I didn't respect the rules
this is my actual progression

 

[Steven G]

I'm sorry I didn't respect the rules
this is my actual progression

Please look at your last post closely.

 

[akramzine]

I'm sorry it's not visible enough i will edit it

 

[akramzine]

I edited my progression