-0.000 000 000 742 147 675 89 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 89(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
-0.000 000 000 742 147 675 89(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:

|-0.000 000 000 742 147 675 89| = 0.000 000 000 742 147 675 89


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: 0.000 000 000 742 147 675 89.

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) 0.000 000 000 742 147 675 89 × 2 = 0 + 0.000 000 001 484 295 351 78;
  • 2) 0.000 000 001 484 295 351 78 × 2 = 0 + 0.000 000 002 968 590 703 56;
  • 3) 0.000 000 002 968 590 703 56 × 2 = 0 + 0.000 000 005 937 181 407 12;
  • 4) 0.000 000 005 937 181 407 12 × 2 = 0 + 0.000 000 011 874 362 814 24;
  • 5) 0.000 000 011 874 362 814 24 × 2 = 0 + 0.000 000 023 748 725 628 48;
  • 6) 0.000 000 023 748 725 628 48 × 2 = 0 + 0.000 000 047 497 451 256 96;
  • 7) 0.000 000 047 497 451 256 96 × 2 = 0 + 0.000 000 094 994 902 513 92;
  • 8) 0.000 000 094 994 902 513 92 × 2 = 0 + 0.000 000 189 989 805 027 84;
  • 9) 0.000 000 189 989 805 027 84 × 2 = 0 + 0.000 000 379 979 610 055 68;
  • 10) 0.000 000 379 979 610 055 68 × 2 = 0 + 0.000 000 759 959 220 111 36;
  • 11) 0.000 000 759 959 220 111 36 × 2 = 0 + 0.000 001 519 918 440 222 72;
  • 12) 0.000 001 519 918 440 222 72 × 2 = 0 + 0.000 003 039 836 880 445 44;
  • 13) 0.000 003 039 836 880 445 44 × 2 = 0 + 0.000 006 079 673 760 890 88;
  • 14) 0.000 006 079 673 760 890 88 × 2 = 0 + 0.000 012 159 347 521 781 76;
  • 15) 0.000 012 159 347 521 781 76 × 2 = 0 + 0.000 024 318 695 043 563 52;
  • 16) 0.000 024 318 695 043 563 52 × 2 = 0 + 0.000 048 637 390 087 127 04;
  • 17) 0.000 048 637 390 087 127 04 × 2 = 0 + 0.000 097 274 780 174 254 08;
  • 18) 0.000 097 274 780 174 254 08 × 2 = 0 + 0.000 194 549 560 348 508 16;
  • 19) 0.000 194 549 560 348 508 16 × 2 = 0 + 0.000 389 099 120 697 016 32;
  • 20) 0.000 389 099 120 697 016 32 × 2 = 0 + 0.000 778 198 241 394 032 64;
  • 21) 0.000 778 198 241 394 032 64 × 2 = 0 + 0.001 556 396 482 788 065 28;
  • 22) 0.001 556 396 482 788 065 28 × 2 = 0 + 0.003 112 792 965 576 130 56;
  • 23) 0.003 112 792 965 576 130 56 × 2 = 0 + 0.006 225 585 931 152 261 12;
  • 24) 0.006 225 585 931 152 261 12 × 2 = 0 + 0.012 451 171 862 304 522 24;
  • 25) 0.012 451 171 862 304 522 24 × 2 = 0 + 0.024 902 343 724 609 044 48;
  • 26) 0.024 902 343 724 609 044 48 × 2 = 0 + 0.049 804 687 449 218 088 96;
  • 27) 0.049 804 687 449 218 088 96 × 2 = 0 + 0.099 609 374 898 436 177 92;
  • 28) 0.099 609 374 898 436 177 92 × 2 = 0 + 0.199 218 749 796 872 355 84;
  • 29) 0.199 218 749 796 872 355 84 × 2 = 0 + 0.398 437 499 593 744 711 68;
  • 30) 0.398 437 499 593 744 711 68 × 2 = 0 + 0.796 874 999 187 489 423 36;
  • 31) 0.796 874 999 187 489 423 36 × 2 = 1 + 0.593 749 998 374 978 846 72;
  • 32) 0.593 749 998 374 978 846 72 × 2 = 1 + 0.187 499 996 749 957 693 44;
  • 33) 0.187 499 996 749 957 693 44 × 2 = 0 + 0.374 999 993 499 915 386 88;
  • 34) 0.374 999 993 499 915 386 88 × 2 = 0 + 0.749 999 986 999 830 773 76;
  • 35) 0.749 999 986 999 830 773 76 × 2 = 1 + 0.499 999 973 999 661 547 52;
  • 36) 0.499 999 973 999 661 547 52 × 2 = 0 + 0.999 999 947 999 323 095 04;
  • 37) 0.999 999 947 999 323 095 04 × 2 = 1 + 0.999 999 895 998 646 190 08;
  • 38) 0.999 999 895 998 646 190 08 × 2 = 1 + 0.999 999 791 997 292 380 16;
  • 39) 0.999 999 791 997 292 380 16 × 2 = 1 + 0.999 999 583 994 584 760 32;
  • 40) 0.999 999 583 994 584 760 32 × 2 = 1 + 0.999 999 167 989 169 520 64;
  • 41) 0.999 999 167 989 169 520 64 × 2 = 1 + 0.999 998 335 978 339 041 28;
  • 42) 0.999 998 335 978 339 041 28 × 2 = 1 + 0.999 996 671 956 678 082 56;
  • 43) 0.999 996 671 956 678 082 56 × 2 = 1 + 0.999 993 343 913 356 165 12;
  • 44) 0.999 993 343 913 356 165 12 × 2 = 1 + 0.999 986 687 826 712 330 24;
  • 45) 0.999 986 687 826 712 330 24 × 2 = 1 + 0.999 973 375 653 424 660 48;
  • 46) 0.999 973 375 653 424 660 48 × 2 = 1 + 0.999 946 751 306 849 320 96;
  • 47) 0.999 946 751 306 849 320 96 × 2 = 1 + 0.999 893 502 613 698 641 92;
  • 48) 0.999 893 502 613 698 641 92 × 2 = 1 + 0.999 787 005 227 397 283 84;
  • 49) 0.999 787 005 227 397 283 84 × 2 = 1 + 0.999 574 010 454 794 567 68;
  • 50) 0.999 574 010 454 794 567 68 × 2 = 1 + 0.999 148 020 909 589 135 36;
  • 51) 0.999 148 020 909 589 135 36 × 2 = 1 + 0.998 296 041 819 178 270 72;
  • 52) 0.998 296 041 819 178 270 72 × 2 = 1 + 0.996 592 083 638 356 541 44;
  • 53) 0.996 592 083 638 356 541 44 × 2 = 1 + 0.993 184 167 276 713 082 88;
  • 54) 0.993 184 167 276 713 082 88 × 2 = 1 + 0.986 368 334 553 426 165 76;

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:


0.000 000 000 742 147 675 89(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 89(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:


0.000 000 000 742 147 675 89(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 -0.000 000 000 742 147 675 89 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