Question Number 208034 by ali009 last updated on 02/Jun/24
Commented by ali009 last updated on 02/Jun/24
$${how}\:{that}\:{can}\:{be}\:{proved}? \\ $$
Answered by Frix last updated on 02/Jun/24
$${s}^{\mathrm{2}} +\frac{{s}}{\mathrm{2}}+\mathrm{2}=\left({s}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{31}}{\mathrm{16}}=\left({s}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}\neq\mathrm{1}.\mathrm{39} \\ $$$$\left({s}+.\mathrm{25}\right)^{\mathrm{2}} +\mathrm{1}.\mathrm{39}^{\mathrm{2}} ={s}^{\mathrm{2}} +.\mathrm{5}{s}+\mathrm{1}.\mathrm{9946} \\ $$$$\mathrm{1}.\mathrm{9946}\neq\mathrm{2} \\ $$$$\mathrm{So}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{proof}. \\ $$
Commented by ali009 last updated on 02/Jun/24
$${i}\:{think}\:{it}'{s} \\ $$$$\sqrt{\mathrm{2}−\mathrm{0}.\mathrm{0625}}\simeq\mathrm{1}.\mathrm{39} \\ $$$${because} \\ $$$$\left({s}+\mathrm{0}.\mathrm{25}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}−\mathrm{0}.\mathrm{0625}}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} +\frac{{s}}{\mathrm{2}}+\mathrm{2} \\ $$
Commented by Frix last updated on 03/Jun/24
$$\mathrm{Mathematical}\:\mathrm{proofs}\:\mathrm{are}\:\mathrm{precise}.\:\mathrm{You}\:\mathrm{can} \\ $$$$\mathrm{round}\:\mathrm{numbers}\:\mathrm{but}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{2}=\mathrm{1}.\mathrm{9946}\:\mathrm{or}\:\frac{\sqrt{\mathrm{31}}}{\mathrm{4}}=\mathrm{1}.\mathrm{39} \\ $$