-0.000 000 000 000 000 000 257 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 257 5(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 000 000 257 5(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 000 000 257 5| = 0.000 000 000 000 000 000 257 5


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 000 257 5.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 000 000 000 257 5 × 2 = 0 + 0.000 000 000 000 000 000 515;
  • 2) 0.000 000 000 000 000 000 515 × 2 = 0 + 0.000 000 000 000 000 001 03;
  • 3) 0.000 000 000 000 000 001 03 × 2 = 0 + 0.000 000 000 000 000 002 06;
  • 4) 0.000 000 000 000 000 002 06 × 2 = 0 + 0.000 000 000 000 000 004 12;
  • 5) 0.000 000 000 000 000 004 12 × 2 = 0 + 0.000 000 000 000 000 008 24;
  • 6) 0.000 000 000 000 000 008 24 × 2 = 0 + 0.000 000 000 000 000 016 48;
  • 7) 0.000 000 000 000 000 016 48 × 2 = 0 + 0.000 000 000 000 000 032 96;
  • 8) 0.000 000 000 000 000 032 96 × 2 = 0 + 0.000 000 000 000 000 065 92;
  • 9) 0.000 000 000 000 000 065 92 × 2 = 0 + 0.000 000 000 000 000 131 84;
  • 10) 0.000 000 000 000 000 131 84 × 2 = 0 + 0.000 000 000 000 000 263 68;
  • 11) 0.000 000 000 000 000 263 68 × 2 = 0 + 0.000 000 000 000 000 527 36;
  • 12) 0.000 000 000 000 000 527 36 × 2 = 0 + 0.000 000 000 000 001 054 72;
  • 13) 0.000 000 000 000 001 054 72 × 2 = 0 + 0.000 000 000 000 002 109 44;
  • 14) 0.000 000 000 000 002 109 44 × 2 = 0 + 0.000 000 000 000 004 218 88;
  • 15) 0.000 000 000 000 004 218 88 × 2 = 0 + 0.000 000 000 000 008 437 76;
  • 16) 0.000 000 000 000 008 437 76 × 2 = 0 + 0.000 000 000 000 016 875 52;
  • 17) 0.000 000 000 000 016 875 52 × 2 = 0 + 0.000 000 000 000 033 751 04;
  • 18) 0.000 000 000 000 033 751 04 × 2 = 0 + 0.000 000 000 000 067 502 08;
  • 19) 0.000 000 000 000 067 502 08 × 2 = 0 + 0.000 000 000 000 135 004 16;
  • 20) 0.000 000 000 000 135 004 16 × 2 = 0 + 0.000 000 000 000 270 008 32;
  • 21) 0.000 000 000 000 270 008 32 × 2 = 0 + 0.000 000 000 000 540 016 64;
  • 22) 0.000 000 000 000 540 016 64 × 2 = 0 + 0.000 000 000 001 080 033 28;
  • 23) 0.000 000 000 001 080 033 28 × 2 = 0 + 0.000 000 000 002 160 066 56;
  • 24) 0.000 000 000 002 160 066 56 × 2 = 0 + 0.000 000 000 004 320 133 12;
  • 25) 0.000 000 000 004 320 133 12 × 2 = 0 + 0.000 000 000 008 640 266 24;
  • 26) 0.000 000 000 008 640 266 24 × 2 = 0 + 0.000 000 000 017 280 532 48;
  • 27) 0.000 000 000 017 280 532 48 × 2 = 0 + 0.000 000 000 034 561 064 96;
  • 28) 0.000 000 000 034 561 064 96 × 2 = 0 + 0.000 000 000 069 122 129 92;
  • 29) 0.000 000 000 069 122 129 92 × 2 = 0 + 0.000 000 000 138 244 259 84;
  • 30) 0.000 000 000 138 244 259 84 × 2 = 0 + 0.000 000 000 276 488 519 68;
  • 31) 0.000 000 000 276 488 519 68 × 2 = 0 + 0.000 000 000 552 977 039 36;
  • 32) 0.000 000 000 552 977 039 36 × 2 = 0 + 0.000 000 001 105 954 078 72;
  • 33) 0.000 000 001 105 954 078 72 × 2 = 0 + 0.000 000 002 211 908 157 44;
  • 34) 0.000 000 002 211 908 157 44 × 2 = 0 + 0.000 000 004 423 816 314 88;
  • 35) 0.000 000 004 423 816 314 88 × 2 = 0 + 0.000 000 008 847 632 629 76;
  • 36) 0.000 000 008 847 632 629 76 × 2 = 0 + 0.000 000 017 695 265 259 52;
  • 37) 0.000 000 017 695 265 259 52 × 2 = 0 + 0.000 000 035 390 530 519 04;
  • 38) 0.000 000 035 390 530 519 04 × 2 = 0 + 0.000 000 070 781 061 038 08;
  • 39) 0.000 000 070 781 061 038 08 × 2 = 0 + 0.000 000 141 562 122 076 16;
  • 40) 0.000 000 141 562 122 076 16 × 2 = 0 + 0.000 000 283 124 244 152 32;
  • 41) 0.000 000 283 124 244 152 32 × 2 = 0 + 0.000 000 566 248 488 304 64;
  • 42) 0.000 000 566 248 488 304 64 × 2 = 0 + 0.000 001 132 496 976 609 28;
  • 43) 0.000 001 132 496 976 609 28 × 2 = 0 + 0.000 002 264 993 953 218 56;
  • 44) 0.000 002 264 993 953 218 56 × 2 = 0 + 0.000 004 529 987 906 437 12;
  • 45) 0.000 004 529 987 906 437 12 × 2 = 0 + 0.000 009 059 975 812 874 24;
  • 46) 0.000 009 059 975 812 874 24 × 2 = 0 + 0.000 018 119 951 625 748 48;
  • 47) 0.000 018 119 951 625 748 48 × 2 = 0 + 0.000 036 239 903 251 496 96;
  • 48) 0.000 036 239 903 251 496 96 × 2 = 0 + 0.000 072 479 806 502 993 92;
  • 49) 0.000 072 479 806 502 993 92 × 2 = 0 + 0.000 144 959 613 005 987 84;
  • 50) 0.000 144 959 613 005 987 84 × 2 = 0 + 0.000 289 919 226 011 975 68;
  • 51) 0.000 289 919 226 011 975 68 × 2 = 0 + 0.000 579 838 452 023 951 36;
  • 52) 0.000 579 838 452 023 951 36 × 2 = 0 + 0.001 159 676 904 047 902 72;
  • 53) 0.001 159 676 904 047 902 72 × 2 = 0 + 0.002 319 353 808 095 805 44;
  • 54) 0.002 319 353 808 095 805 44 × 2 = 0 + 0.004 638 707 616 191 610 88;
  • 55) 0.004 638 707 616 191 610 88 × 2 = 0 + 0.009 277 415 232 383 221 76;
  • 56) 0.009 277 415 232 383 221 76 × 2 = 0 + 0.018 554 830 464 766 443 52;
  • 57) 0.018 554 830 464 766 443 52 × 2 = 0 + 0.037 109 660 929 532 887 04;
  • 58) 0.037 109 660 929 532 887 04 × 2 = 0 + 0.074 219 321 859 065 774 08;
  • 59) 0.074 219 321 859 065 774 08 × 2 = 0 + 0.148 438 643 718 131 548 16;
  • 60) 0.148 438 643 718 131 548 16 × 2 = 0 + 0.296 877 287 436 263 096 32;
  • 61) 0.296 877 287 436 263 096 32 × 2 = 0 + 0.593 754 574 872 526 192 64;
  • 62) 0.593 754 574 872 526 192 64 × 2 = 1 + 0.187 509 149 745 052 385 28;
  • 63) 0.187 509 149 745 052 385 28 × 2 = 0 + 0.375 018 299 490 104 770 56;
  • 64) 0.375 018 299 490 104 770 56 × 2 = 0 + 0.750 036 598 980 209 541 12;
  • 65) 0.750 036 598 980 209 541 12 × 2 = 1 + 0.500 073 197 960 419 082 24;
  • 66) 0.500 073 197 960 419 082 24 × 2 = 1 + 0.000 146 395 920 838 164 48;
  • 67) 0.000 146 395 920 838 164 48 × 2 = 0 + 0.000 292 791 841 676 328 96;
  • 68) 0.000 292 791 841 676 328 96 × 2 = 0 + 0.000 585 583 683 352 657 92;
  • 69) 0.000 585 583 683 352 657 92 × 2 = 0 + 0.001 171 167 366 705 315 84;
  • 70) 0.001 171 167 366 705 315 84 × 2 = 0 + 0.002 342 334 733 410 631 68;
  • 71) 0.002 342 334 733 410 631 68 × 2 = 0 + 0.004 684 669 466 821 263 36;
  • 72) 0.004 684 669 466 821 263 36 × 2 = 0 + 0.009 369 338 933 642 526 72;
  • 73) 0.009 369 338 933 642 526 72 × 2 = 0 + 0.018 738 677 867 285 053 44;
  • 74) 0.018 738 677 867 285 053 44 × 2 = 0 + 0.037 477 355 734 570 106 88;
  • 75) 0.037 477 355 734 570 106 88 × 2 = 0 + 0.074 954 711 469 140 213 76;
  • 76) 0.074 954 711 469 140 213 76 × 2 = 0 + 0.149 909 422 938 280 427 52;
  • 77) 0.149 909 422 938 280 427 52 × 2 = 0 + 0.299 818 845 876 560 855 04;
  • 78) 0.299 818 845 876 560 855 04 × 2 = 0 + 0.599 637 691 753 121 710 08;
  • 79) 0.599 637 691 753 121 710 08 × 2 = 1 + 0.199 275 383 506 243 420 16;
  • 80) 0.199 275 383 506 243 420 16 × 2 = 0 + 0.398 550 767 012 486 840 32;
  • 81) 0.398 550 767 012 486 840 32 × 2 = 0 + 0.797 101 534 024 973 680 64;
  • 82) 0.797 101 534 024 973 680 64 × 2 = 1 + 0.594 203 068 049 947 361 28;
  • 83) 0.594 203 068 049 947 361 28 × 2 = 1 + 0.188 406 136 099 894 722 56;
  • 84) 0.188 406 136 099 894 722 56 × 2 = 0 + 0.376 812 272 199 789 445 12;
  • 85) 0.376 812 272 199 789 445 12 × 2 = 0 + 0.753 624 544 399 578 890 24;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 000 000 000 257 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 0000 0000 0010 0110 0(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 257 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 0000 0000 0010 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 62 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 257 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 0000 0000 0010 0110 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1100 0000 0000 0010 0110 0(2) × 20 =

1.0011 0000 0000 0000 1001 100(2) × 2-62


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -62

Mantissa (not normalized):
1.0011 0000 0000 0000 1001 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-62 + 2(8-1) - 1 =

(-62 + 127)(10) =

65(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

65(10) =

0100 0001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 001 1000 0000 0000 0100 1100 =

001 1000 0000 0000 0100 1100


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0100 0001

Mantissa (23 bits) =
001 1000 0000 0000 0100 1100


Decimal number -0.000 000 000 000 000 000 257 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0100 0001 - 001 1000 0000 0000 0100 1100

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111