Thread: Please need help with direct proof

Hi, I have a few problems which i can't solve

1) need to prove (m+dk)mod d = m mod d (where m, d and k are integers and d does not equal to zero)

2) if r is any rational number and s is any irrational, then r/s is irrational

3) prove that there is at most one real number b with the property that br = r for all real numbers r.

Thank you

Hello, somestudent2!

Prove: .
where are integers and .

Definition: .

In baby-talk: and are congruent (mod ) if their difference is a multiple of


Since , a multiple of
. . then: .

2) If is any rational number and is any irrational, then is irrational.

Proof by contradiction: .Assume is rational.

Then: . .for some integers and [1]

Since is rational: . for some integers and

Then equation [1] becomes: .


Since and are integers (integers are closed under multiplication),
. . is a rational number. .But we are told that is irrational.

We have reached a contradiction, hence our assumption is incorrect.

. . Therefore: . is irrational.

[Oh-oh! .You said direct proof, didn't you? . . . Never mind.]

Originally Posted by somestudent2

3) prove that there is at most one real number b with the property that br = r for all real numbers r.

Say that some other real number, e, also has the property that er = r for all real numbers r. Then






Thus b = e or r = 0. But r is any real number, so r is not 0 in general. Thus b = e.

-Dan

Originally Posted by topsquark

But r is any real number, so r is not 0 in general.

But zero is a real number, you cannot conclude it is non-zero. So his statement is wrong as you have shown.

Thanks a lot for the quick response guys. I actually managed to solve 2nd and 3rd problem this morning today, and was glad to find out that my solution for 2nd matched with Soroban's solution(although I used square root of 2 to represent irrational number)

For the last problem a sol. through contradiction came to my mind
First we negate the statement, so we have
Suppose there are two distinct real numbers a1 and a2 such that
1) a1*r = r 2) a2*r = r

then we have a1*a2= a2 (if r=a2)
and a2*a1=a1 (if r=a1)
this emplies

a2=a1*a2=a2*a1= a1 (what means that a2=a1), but we assumed that a1 and a2 are two distinct real numbers, so we have a contradiction and thus our assumption was wrong.

please tell me if this proof works.

2)Is there a way to prove m+dk mod d = m mod d (without the use of the definition of mod, I don't think I can use it just yet)

I was trying to get it going

by representing m = dq +r and m+dk = dp +r
then m=d(p-k) + r
how do I prove that rs are the same in both cases?

Thank you