Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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What are the philosophers paid for?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

For one thing they can head over to a university and find teaching jobs or author a book.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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It looks like they are teaching each other expressions.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

They are both going Hmmm.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

bob bundy wrote:

To extract a contradiction argument you would have to assume that x + y ≠ 88 and show this leads to a contradiction. Since many other values (other than 88) are possible, I think you'd have a tough time with this.

Bob

It might be doable -- I haven't tried via contradiction, although I'd assume you'd start with *x + y < 88* and find bounds for *a + b + ab* that don't include 2020, then do something similar for the case *x + y > 88*. The AM-GM inequality might help for this due to the ab term. However, the problem seems engineered to make use of the factorisation *(a + 1)(b + 1) - 1* = *a + b + ab*.