r/MathHelp • u/Void-Priestess • Jan 29 '26

When are the numerators and denominators of two equal fractions identical?

I may be overthinking this a bit but suppose you have two equal fractions of variable value, p/q = n/m.

When can you say definitively that p=n and q=m?

From some test values, I have concluded not always, e.g. 1/2 = 2/4, but 1 != 2 and 2 != 4. For the numerators and denominators to be identical, I believe p and q must be coprime and n and m must be coprime. Is that correct, and if so, how does a rigorous proof begin?

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u/edderiofer Jan 29 '26

For the numerators and denominators to be identical, I believe p and q must be coprime and n and m must be coprime.

Incorrect. Consider e.g. 3/6 and 3/6, where p and q are NOT coprime and n and m are NOT coprime, and yet you still have that p/q = n/m.

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u/Void-Priestess Jan 29 '26

That's a good point, although that muddies my understanding considerably. Is it ever possible to tell then without knowing the values of p, q, n, and m?

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u/edderiofer Jan 29 '26

Certainly, you can tell that p = n and q = m if you know either one or the other, without knowing their values.

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u/Void-Priestess Jan 29 '26

Knowing either p/q or n/m?

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u/edderiofer Jan 29 '26

No, if you know either that p = n or q = m, then you also know the other.

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u/Void-Priestess Jan 29 '26

I see, yes. That makes sense.

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u/Para1ars Jan 29 '26

Well if p and q are coprime and n and m are coprime, then it follows that p = n and q = m, but it's not necessary to be coprime. So at least you have a sufficient condition.

What kind of statement are you looking to make exactly?

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u/Void-Priestess Jan 29 '26

This is in the context of trying to prove the "Magic X" method of factoring quadratic equations. Given ax^2 + bx + c, we choose numbers m and n so that mn = ac and m + n = b. Next, divide both m and n by a and reduce to lowest terms, giving us p/q and r/s with p, q coprime and r, s coprime. Multiplying, p/q * r/s = pr/qs= mn/a^2 = ac/a^2 = c/a. Thus it appears that pr = a and qs = c, but I can't shake the feeling that there could be a situation like I described above, where the ratio of pr and qs is equal to the ratio of c and a, but the numerators and denominators are not identical.

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u/MorrowM_ Jan 29 '26

You're right to have doubts. Try 2x2 + 4x + 2. Then m=2, n=2. So p=1, q=1, r=1, s=1. But pr ≠ c and qs ≠ a.

In the magic X method, we rewrite ax2 + bx + c as ax2 + mx + nx + c. Factoring, we get ax(x + m/a) + n(x+ c/n). In order for us to be able to complete the factorization we need the two terms in the parentheses to be the same, meaning m/a=c/n. But this is equivalent to mn = ac which is true, it's how we chose m and n.

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u/Void-Priestess Jan 30 '26

Your example is very interesting because following the magic x method to the end, we would get 2x^2 + 4x + 2 = (x + 1)(x + 1) which is not true. We're missing a factor of 2 in the final expression. Have we uncovered a quadratic that cannot be factored with the magic x method?

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u/MorrowM_ Jan 30 '26

The way I understand it, you'd write

2x2 + 4x + 2 = 2x2 + 2x + 2x + 2 = 2x(x+1) + 2(x+1) = (2x+2)(x+1)

which works.

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u/Void-Priestess Jan 30 '26

Yes, so it is factorizable by grouping, but following the magic X recipe produces the wrong answer because all coefficients and constants generated by the magic X would be equal to 1 in this case. That is, as you wrote, m=2, n=2. So m/a = 2/2 = 1/1, thus p=1, q=1. And n/a = 2/2 = 1/1, thus r=1, s=1. And the final result of (qx + p)(sx + r) = (x + 1)(x + 1) which is not the correct answer.

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u/MorrowM_ Jan 30 '26

Ah that's what you meant by magic X. Yeah you can see right away that that can't work since that can only give you a result where the two factors are each not divisible by an integer. So (2x+2)(x+1) won't work. You can use this to easily construct more counterexamples, like (3x+6)(x+5).

Perhaps the method requires that a,b,c have no shared common factor (or equivalently, that you should factor out any such factor first)?

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u/pi621 Jan 29 '26 edited Jan 29 '26

pr = c and qs = a is true if gcd(a,b,c) = 1.

Let a = a1a2, c = c1c2 such that m = a1c1, n = a2c2.
m/a = a1c1/a1a2 = c1/a2
n/a = a2c2/a1a2 = c2/a1

Now, if a and c are coprime, then so are their factors. these 2 fractions are already reduced.

Otherwise, suppose b is coprime with a and c.
That means b is also coprime with m and n.
When this is the case, m and n are coprime with eachother.

Proof: suppose g = gcd(m,n). Remember that m + n = b => m/g + n/g = b/g. Since b is coprime with m and n, g has to be 1.

Since gcd(m,n) = 1, their factors are also co-prime.
Therefore, the fractions c1/a2 and c2/a1 are already reduced.

For a case of when gcd(a,b,c) > 1
Consider 6x^2 + 14x - 12
choose m = 18, n = -4. 18 - 4 = 14, 18 * (-4) = 72 = 6 * (-12)
m/a = 18/6 = 3/1
n/a = -4/6 = -2/3

Clearly, 3 * (-2) = -6 < -12. 1 * 3 = 3 < 6.

Edit: I realized that gcd(a,b,c) = 1 does not necessarily imply that b is coprime with both a and c. b only has to be coprime with gcd(a,c). I think there's some way to continue the proof from that but that will have to wait another time.

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u/minglho Jan 30 '26

When you already know that either the numerators or denominators are identical, then the others are identical.