-0.000 000 000 742 147 676 58 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 58(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
-0.000 000 000 742 147 676 58(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:

|-0.000 000 000 742 147 676 58| = 0.000 000 000 742 147 676 58


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: 0.000 000 000 742 147 676 58.

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) 0.000 000 000 742 147 676 58 × 2 = 0 + 0.000 000 001 484 295 353 16;
  • 2) 0.000 000 001 484 295 353 16 × 2 = 0 + 0.000 000 002 968 590 706 32;
  • 3) 0.000 000 002 968 590 706 32 × 2 = 0 + 0.000 000 005 937 181 412 64;
  • 4) 0.000 000 005 937 181 412 64 × 2 = 0 + 0.000 000 011 874 362 825 28;
  • 5) 0.000 000 011 874 362 825 28 × 2 = 0 + 0.000 000 023 748 725 650 56;
  • 6) 0.000 000 023 748 725 650 56 × 2 = 0 + 0.000 000 047 497 451 301 12;
  • 7) 0.000 000 047 497 451 301 12 × 2 = 0 + 0.000 000 094 994 902 602 24;
  • 8) 0.000 000 094 994 902 602 24 × 2 = 0 + 0.000 000 189 989 805 204 48;
  • 9) 0.000 000 189 989 805 204 48 × 2 = 0 + 0.000 000 379 979 610 408 96;
  • 10) 0.000 000 379 979 610 408 96 × 2 = 0 + 0.000 000 759 959 220 817 92;
  • 11) 0.000 000 759 959 220 817 92 × 2 = 0 + 0.000 001 519 918 441 635 84;
  • 12) 0.000 001 519 918 441 635 84 × 2 = 0 + 0.000 003 039 836 883 271 68;
  • 13) 0.000 003 039 836 883 271 68 × 2 = 0 + 0.000 006 079 673 766 543 36;
  • 14) 0.000 006 079 673 766 543 36 × 2 = 0 + 0.000 012 159 347 533 086 72;
  • 15) 0.000 012 159 347 533 086 72 × 2 = 0 + 0.000 024 318 695 066 173 44;
  • 16) 0.000 024 318 695 066 173 44 × 2 = 0 + 0.000 048 637 390 132 346 88;
  • 17) 0.000 048 637 390 132 346 88 × 2 = 0 + 0.000 097 274 780 264 693 76;
  • 18) 0.000 097 274 780 264 693 76 × 2 = 0 + 0.000 194 549 560 529 387 52;
  • 19) 0.000 194 549 560 529 387 52 × 2 = 0 + 0.000 389 099 121 058 775 04;
  • 20) 0.000 389 099 121 058 775 04 × 2 = 0 + 0.000 778 198 242 117 550 08;
  • 21) 0.000 778 198 242 117 550 08 × 2 = 0 + 0.001 556 396 484 235 100 16;
  • 22) 0.001 556 396 484 235 100 16 × 2 = 0 + 0.003 112 792 968 470 200 32;
  • 23) 0.003 112 792 968 470 200 32 × 2 = 0 + 0.006 225 585 936 940 400 64;
  • 24) 0.006 225 585 936 940 400 64 × 2 = 0 + 0.012 451 171 873 880 801 28;
  • 25) 0.012 451 171 873 880 801 28 × 2 = 0 + 0.024 902 343 747 761 602 56;
  • 26) 0.024 902 343 747 761 602 56 × 2 = 0 + 0.049 804 687 495 523 205 12;
  • 27) 0.049 804 687 495 523 205 12 × 2 = 0 + 0.099 609 374 991 046 410 24;
  • 28) 0.099 609 374 991 046 410 24 × 2 = 0 + 0.199 218 749 982 092 820 48;
  • 29) 0.199 218 749 982 092 820 48 × 2 = 0 + 0.398 437 499 964 185 640 96;
  • 30) 0.398 437 499 964 185 640 96 × 2 = 0 + 0.796 874 999 928 371 281 92;
  • 31) 0.796 874 999 928 371 281 92 × 2 = 1 + 0.593 749 999 856 742 563 84;
  • 32) 0.593 749 999 856 742 563 84 × 2 = 1 + 0.187 499 999 713 485 127 68;
  • 33) 0.187 499 999 713 485 127 68 × 2 = 0 + 0.374 999 999 426 970 255 36;
  • 34) 0.374 999 999 426 970 255 36 × 2 = 0 + 0.749 999 998 853 940 510 72;
  • 35) 0.749 999 998 853 940 510 72 × 2 = 1 + 0.499 999 997 707 881 021 44;
  • 36) 0.499 999 997 707 881 021 44 × 2 = 0 + 0.999 999 995 415 762 042 88;
  • 37) 0.999 999 995 415 762 042 88 × 2 = 1 + 0.999 999 990 831 524 085 76;
  • 38) 0.999 999 990 831 524 085 76 × 2 = 1 + 0.999 999 981 663 048 171 52;
  • 39) 0.999 999 981 663 048 171 52 × 2 = 1 + 0.999 999 963 326 096 343 04;
  • 40) 0.999 999 963 326 096 343 04 × 2 = 1 + 0.999 999 926 652 192 686 08;
  • 41) 0.999 999 926 652 192 686 08 × 2 = 1 + 0.999 999 853 304 385 372 16;
  • 42) 0.999 999 853 304 385 372 16 × 2 = 1 + 0.999 999 706 608 770 744 32;
  • 43) 0.999 999 706 608 770 744 32 × 2 = 1 + 0.999 999 413 217 541 488 64;
  • 44) 0.999 999 413 217 541 488 64 × 2 = 1 + 0.999 998 826 435 082 977 28;
  • 45) 0.999 998 826 435 082 977 28 × 2 = 1 + 0.999 997 652 870 165 954 56;
  • 46) 0.999 997 652 870 165 954 56 × 2 = 1 + 0.999 995 305 740 331 909 12;
  • 47) 0.999 995 305 740 331 909 12 × 2 = 1 + 0.999 990 611 480 663 818 24;
  • 48) 0.999 990 611 480 663 818 24 × 2 = 1 + 0.999 981 222 961 327 636 48;
  • 49) 0.999 981 222 961 327 636 48 × 2 = 1 + 0.999 962 445 922 655 272 96;
  • 50) 0.999 962 445 922 655 272 96 × 2 = 1 + 0.999 924 891 845 310 545 92;
  • 51) 0.999 924 891 845 310 545 92 × 2 = 1 + 0.999 849 783 690 621 091 84;
  • 52) 0.999 849 783 690 621 091 84 × 2 = 1 + 0.999 699 567 381 242 183 68;
  • 53) 0.999 699 567 381 242 183 68 × 2 = 1 + 0.999 399 134 762 484 367 36;
  • 54) 0.999 399 134 762 484 367 36 × 2 = 1 + 0.998 798 269 524 968 734 72;

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:


0.000 000 000 742 147 676 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 58 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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