# Thread: Please need help with direct proof

1. ## 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

2. Hello, somestudent2!

Prove: .$\displaystyle m + dk \equiv m \pmod{d}$
where $\displaystyle m,\,d,\,k$ are integers and $\displaystyle d \neq 0$.

Definition: .$\displaystyle a \equiv b \pmod {d} \; \Longleftrightarrow \;a - b\:=\:dn$

In baby-talk: $\displaystyle a$ and $\displaystyle b$ are congruent (mod $\displaystyle d$) if their difference is a multiple of $\displaystyle d.$

Since $\displaystyle (m + dk) - m \:=\:dk$, a multiple of $\displaystyle d,$
. . then: .$\displaystyle m + dk \equiv m \pmod{d}$

2) If $\displaystyle r$ is any rational number and $\displaystyle s$ is any irrational, then $\displaystyle \frac{r}{s}$ is irrational.

Proof by contradiction: .Assume $\displaystyle \frac{r}{s}$ is rational.

Then: .$\displaystyle \frac{r}{s}\,=\,\frac{a}{b}$ .for some integers $\displaystyle a$ and $\displaystyle b.$ [1]

Since $\displaystyle r$ is rational: .$\displaystyle r \,=\,\frac{c}{d}$ for some integers $\displaystyle c$ and $\displaystyle d.$

Then equation [1] becomes: .$\displaystyle \frac{\frac{c}{d}}{s} \,=\,\frac{a}{b}\quad\Rightarrow\quad s \,=\,\frac{bc}{ad}$

Since $\displaystyle bc$ and $\displaystyle ad$ are integers (integers are closed under multiplication),
. . $\displaystyle s$ is a rational number. .But we are told that $\displaystyle s$ is irrational.

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

. . Therefore: .$\displaystyle \frac{r}{s}$ is irrational.

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

3. 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
$\displaystyle br = r = er$

$\displaystyle br - er = 0$

$\displaystyle (b - e)r = 0$

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

-Dan

4. 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.

5. 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