I'm learning the absolute basics of how to do proofs, and am really struggling.

If 3x is even then 3x+5 is odd.

This is the solution:

enter image description here

I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb. I tried looking up other answers before posting, but nothing I found is this basic.

Work: -Assumptions- 3x = 2n 3x+5 = 2k+1

-Trying to make sense of 3x- 3x+5 = 2k+1 3x = 2k-4

-Plugging in 2k-4 for 3x- 2k-4 = 2n 2k = 2n+4 k = n+2

-Plugging in n+2 for k- 3x+5 = 2(n+2)+1

...This is where I gave up. I don't know where I'm going with this anymore.

  • 4
    $\begingroup$ You should calm down, everyone can have difficulties. Explain us where you get lost. If you don't know where you get lost I can try to explain you the proof step by step. $\endgroup$ – Eureka Dec 5 '20 at 17:42
  • $\begingroup$ I'm good, just frustrated. I'd like step by step. I'll edit in my work above. Warning, there is no rhyme or reason to it. $\endgroup$ – visualbread Dec 5 '20 at 17:44
  • 2
    $\begingroup$ If $y$ is even, can you prove that $y+5$ is odd? Try this first, then try your problem. There’s no substantial difference. $\endgroup$ – Benjamin Wang Dec 5 '20 at 17:46
  • 1
    $\begingroup$ @visualbread your resolution may be a bit convoluted but it's correct. You just proved that $3x+5=2(n+2)+1$. And $2(n+2)+1$ is odd ,because it is of the form $2h+1$ with $h=n+2$ $\endgroup$ – Eureka Dec 5 '20 at 17:52
  • $\begingroup$ @BenjaminWang I am unable to do that one either. From y = 2n and y+5 = 2k+1, I can get that all the way to n = k -2 (or k = n+2), but I can't see what that tells me. The final solution has everything nicely equal to one another, and I can't seem to set it up that way. $\endgroup$ – visualbread Dec 5 '20 at 17:54

I think you're getting confused by the $3x$ part. The $3x$ plays no role in the problem.

Suppose you're given any number that is even. I'll call it y. Now we want to show $y+5$ is odd.

Therefore by definition

$y=2k$ for some integer $k$


$y+5 = 2k+5$

Now we just need to show that $2k+5$ is 2 times an integer plus 1

$2k+5 = 2(k+2)+1$

So $2k+5$ is odd because it can be written in the form 2*integer +1 where the integer here is $k+2$. So $y+5$ is odd since $y+5 = 2k+5$

So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...

  • $\begingroup$ I might be there, but I need just a little bit more help. How did you write 2k+5 in the form 2(k+2) + 1? I can see that those are the same, but how could I do it without trial and error? $\endgroup$ – visualbread Dec 5 '20 at 18:16
  • 2
    $\begingroup$ So the first step is to look at 5. We see it is an odd integer. So I decide it is more convenient to write it as $4+1$. So I write $2k+5$ as $2k+4+1$. Now I can factor $2k+4 = 2(k+2)$. So putting it all together $2k+5 = 2k+4+1 = 2(k+2)+1$. Hope this makes sense. let me know. Suppose I had $2k+6$ instead. since 6 is even, I know I can factor immediately $2k+6 = 2(k+3)$. $\endgroup$ – Ameet Sharma Dec 5 '20 at 18:19
  • $\begingroup$ Thanks! This shows me all the steps I need to follow my book's logic. I don't know what happened in my math education, but I never did learn some of these little things that make some of these proofs actually understandable. At least there's websites like this one to fill in the gaps. $\endgroup$ – visualbread Dec 5 '20 at 18:25
  • $\begingroup$ No problem. I highly recommend doing lots of exercises with factoring, simplification and algebra so that these things become second nature. It takes a lot of practice but it is worth it. As you get into more advanced math, these things really need to be 2nd nature where you don't need to think about them anymore... otherwise you'll get bogged down at every step. $\endgroup$ – Ameet Sharma Dec 5 '20 at 18:31
  • 1
    $\begingroup$ I'm sure there are lots of online websites. Here's one I just found as part of an online algebra textbook: openstax.org/books/college-algebra/pages/1-review-exercises. Each chapter has review exercises and a practice test. I recommend doing all these, maybe multiple times till you can do them in your sleep. In my opinion... if you want online resources, look for online algebra "textbooks"... and do all the exercises in each chapter. If you have doubts you can review the chapter. $\endgroup$ – Ameet Sharma Dec 5 '20 at 18:51

Your problem is that $3x + 5 = 2k +1$ is not an assumption. It is the conclusion you need to prove.

Your one and only assumption is that $3x = 2n$ for some integer $n$.

so you start with

$3x = 2n$.

.... then you do a bunch of steps ....

.... steps .....

.... and get in the end ........

Conclusion: $3x + 5 = 2(????????) + 1$ where $??????$ is some integer you come up with in you steps.

Let's see what happens when we try. Let's take it nice and slow:


$3x = 2n$.

$3x +5 = 2n + 5$

....hmmm, we want $2(??????) + \color{red}1$ in the end so let's pull out the $+\color{red}1$ first.....

$3x + 5 = 2n+5 = 2n + (4 + \color{red}1)=(2n+4) +\color{red}1$

.... hmmm, okay that's the $+1$ now we want $2(\color{red}{??????}) + 1$. To get the So we need to factor then $2$ out of $2n+4$ and see what we have left.... that will bee the $\color{red}{??????}$

$3x + 5 = (\color{red}{2n+4}) + 1$

$3x + 5= 2(\color{red}{n + 2}) + 1$

.... and that's it......

Conclusion: $3x+5 = 2(\color{red}{n + 2})+1$.

The $??????$ we wanted turns out to be $\color{red}{n+2}$ an we have

$3x + 5 = (3x+4) + 1 = (2n+4) + 1 = 2(\color{red}{n+2}) + 1$.

And because $\color{red}{n+2}$ is an integer if we let $k = n+2$ be that integer $3x+5 = 2k + 1$ and so... $3x + 5$ is odd.


Although if you want to work backwards

Conclusion: $3x + 5 = 2k +1$ ..... and we want to solve for $k$ to show it is possible...

$3x + 5 -1 =2k + 1-1$

$3x +4 = 2k$

$k = \frac {3x + 4}2 = \frac {3x}2 + 2$.

.... but is $\frac {3x}2 + 2$ an integer?????

Well, $3x$ is even. So there is an integer $n$ so that $3x = 2n$ so

$k = \frac {3x}2 +2 = \frac {2n}2 + 2 = n+2$.

So $k=n+2$ is the integer we want to conclude $3x+5 =2k +1$.

If we did it this way our proof would go:

$3x$ is even so there is an integer $n$ so that $3x = 2n$. Let $k = n+2$; that is an integer.

$2k + 1 = 2(n+2)+1 = 2n + 5 = 3x + 5$.

So $3x+5 = 2k +1$ and that is odd.


Ok so we know that $3x$ is even, that means we can write $3x=2n$ for a suitable $n$, since even means that the number is divisible by two without remainder. But then we have $3x+5=2n+5=2n+(4+1)=2n+2\cdot 2+1=2(n+2)+1$ which clearly is odd.

  • $\begingroup$ Thanks for the reply. I'm still not there yet. I can't figure out where 2n+5 came from or why it can be set equal to 2x+5. I need this broken down even further if that's even possible. $\endgroup$ – visualbread Dec 5 '20 at 17:55
  • $\begingroup$ We set it equal to $3x+5$ NOT $2x+5$. I will try to break it down further: so $\endgroup$ – Mo145 Dec 5 '20 at 18:04
  • $\begingroup$ 1)a natural number $N$ is called even if it is a multiple of $2$. So per definition we can write every even number $N$ as $N=2M$. Note that we can write any natural number as $N=2\frac{N}{2}$ but in general $\frac{N}{2}$ will not be a natural number, for example take $N=3$, then clearly $\frac{3}{2}$ is not a natural number. In fact this will only be the case if $N$ is even. $\endgroup$ – Mo145 Dec 5 '20 at 18:08
  • $\begingroup$ 2)Since we assume $3x$ to be even we can, by 1) write it as $3x=2n$ where $n=\frac{3x}{2}$ so we have the following equality $3x=2n$ adding a number on both sides preserves the equality hence we have $3x+5=2n+5$ $\endgroup$ – Mo145 Dec 5 '20 at 18:11
  • $\begingroup$ I hope this clarifies your concerns :) $\endgroup$ – Mo145 Dec 5 '20 at 18:11

We know that $3x=2n$. So: $$\color{blue}{3x}+5=\color{blue}{2n}+5$$ Because the "blue quantities" are equal. Now: $$3x+5=2n+5=2n+4+1$$ In this step I just wrote $5$ as $4+1$:

$$ 3x+5=2n+5=2n+4+1=2n+2\cdot 2+1$$ In this step I just wrote $4$ as $2 \cdot 2$.

$$ 3x+5=2n+5=2n+4+1=\color{green}{2n+2\cdot 2}+1=\color{green}{2(n+2)}+1$$ The last step is valid because the green quantities are equal. In the end: $$3x+5=2(n+2)+1$$ This means that $3x+5$ is odd because is of the form $2h+1$ with $h$ integer(in particular $h=(n+2)$)


Remember here that $n$ represents ANY natural number. You got to the answer but you didn’t even realize it. That’s probably because you are thinking syntactically rather than semantically. What I mean is the literal string of symbols $2(n+2)$ didn’t register to you as even because it is not the same as the string $2n$. But $n+2$ is a natural number just like $n$ is. So the the strings $2(n+2)$ and $2n$ both represent even numbers, and so $2(n+2) + 1$ is odd just as you have shown in your last line.


You are dealing with a problem where you are given too much information. Here is another way of doing it (we'll use $m$ for an integer).

If $3x$ is even then $x$ must be even, so we can put $x=2m$. [This is a consequence of the fact that $2$ is a prime, or can be proved in various ways]

Then $3x+5=6m+5=6m+4+1=2(3m+2)+1$.

Now we can put $3m=n$, an integer, to get $2(n+2)+1$.

For an alternative proof, we could put $3m+2=n$ and then we get $2n+1$.

Note that the last two sentences are alternatives to one another. They both use $n$, but $n$ is defined differently in the two cases.

What I'm trying to do here is to unpack how the different expressions for the same thing relate to each other. If you get your head round that you will be flying.

  • 1
    $\begingroup$ I find this answer to be pretty overwhelming for the OP, since you are talking about "alternative proofs" and "consequences of the fact that 2 is prime that can be proved in various ways". I'm not underestimating the OP, but I think that when you learn fron basic you should do things calmly step by step. $\endgroup$ – Eureka Dec 5 '20 at 18:23
  • $\begingroup$ @Eureka What I was trying to do was to show what happens when you keep the factor $3$ for an extra step. If it is "overwhelming" it might help OP to understand why the factor $3$ is best dropped ("too much information"). Either it helps or it doesn't. $\endgroup$ – Mark Bennet Dec 5 '20 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.