-0.000 000 000 742 147 674 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 72(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
-0.000 000 000 742 147 674 72(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:

|-0.000 000 000 742 147 674 72| = 0.000 000 000 742 147 674 72


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: 0.000 000 000 742 147 674 72.

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) 0.000 000 000 742 147 674 72 × 2 = 0 + 0.000 000 001 484 295 349 44;
  • 2) 0.000 000 001 484 295 349 44 × 2 = 0 + 0.000 000 002 968 590 698 88;
  • 3) 0.000 000 002 968 590 698 88 × 2 = 0 + 0.000 000 005 937 181 397 76;
  • 4) 0.000 000 005 937 181 397 76 × 2 = 0 + 0.000 000 011 874 362 795 52;
  • 5) 0.000 000 011 874 362 795 52 × 2 = 0 + 0.000 000 023 748 725 591 04;
  • 6) 0.000 000 023 748 725 591 04 × 2 = 0 + 0.000 000 047 497 451 182 08;
  • 7) 0.000 000 047 497 451 182 08 × 2 = 0 + 0.000 000 094 994 902 364 16;
  • 8) 0.000 000 094 994 902 364 16 × 2 = 0 + 0.000 000 189 989 804 728 32;
  • 9) 0.000 000 189 989 804 728 32 × 2 = 0 + 0.000 000 379 979 609 456 64;
  • 10) 0.000 000 379 979 609 456 64 × 2 = 0 + 0.000 000 759 959 218 913 28;
  • 11) 0.000 000 759 959 218 913 28 × 2 = 0 + 0.000 001 519 918 437 826 56;
  • 12) 0.000 001 519 918 437 826 56 × 2 = 0 + 0.000 003 039 836 875 653 12;
  • 13) 0.000 003 039 836 875 653 12 × 2 = 0 + 0.000 006 079 673 751 306 24;
  • 14) 0.000 006 079 673 751 306 24 × 2 = 0 + 0.000 012 159 347 502 612 48;
  • 15) 0.000 012 159 347 502 612 48 × 2 = 0 + 0.000 024 318 695 005 224 96;
  • 16) 0.000 024 318 695 005 224 96 × 2 = 0 + 0.000 048 637 390 010 449 92;
  • 17) 0.000 048 637 390 010 449 92 × 2 = 0 + 0.000 097 274 780 020 899 84;
  • 18) 0.000 097 274 780 020 899 84 × 2 = 0 + 0.000 194 549 560 041 799 68;
  • 19) 0.000 194 549 560 041 799 68 × 2 = 0 + 0.000 389 099 120 083 599 36;
  • 20) 0.000 389 099 120 083 599 36 × 2 = 0 + 0.000 778 198 240 167 198 72;
  • 21) 0.000 778 198 240 167 198 72 × 2 = 0 + 0.001 556 396 480 334 397 44;
  • 22) 0.001 556 396 480 334 397 44 × 2 = 0 + 0.003 112 792 960 668 794 88;
  • 23) 0.003 112 792 960 668 794 88 × 2 = 0 + 0.006 225 585 921 337 589 76;
  • 24) 0.006 225 585 921 337 589 76 × 2 = 0 + 0.012 451 171 842 675 179 52;
  • 25) 0.012 451 171 842 675 179 52 × 2 = 0 + 0.024 902 343 685 350 359 04;
  • 26) 0.024 902 343 685 350 359 04 × 2 = 0 + 0.049 804 687 370 700 718 08;
  • 27) 0.049 804 687 370 700 718 08 × 2 = 0 + 0.099 609 374 741 401 436 16;
  • 28) 0.099 609 374 741 401 436 16 × 2 = 0 + 0.199 218 749 482 802 872 32;
  • 29) 0.199 218 749 482 802 872 32 × 2 = 0 + 0.398 437 498 965 605 744 64;
  • 30) 0.398 437 498 965 605 744 64 × 2 = 0 + 0.796 874 997 931 211 489 28;
  • 31) 0.796 874 997 931 211 489 28 × 2 = 1 + 0.593 749 995 862 422 978 56;
  • 32) 0.593 749 995 862 422 978 56 × 2 = 1 + 0.187 499 991 724 845 957 12;
  • 33) 0.187 499 991 724 845 957 12 × 2 = 0 + 0.374 999 983 449 691 914 24;
  • 34) 0.374 999 983 449 691 914 24 × 2 = 0 + 0.749 999 966 899 383 828 48;
  • 35) 0.749 999 966 899 383 828 48 × 2 = 1 + 0.499 999 933 798 767 656 96;
  • 36) 0.499 999 933 798 767 656 96 × 2 = 0 + 0.999 999 867 597 535 313 92;
  • 37) 0.999 999 867 597 535 313 92 × 2 = 1 + 0.999 999 735 195 070 627 84;
  • 38) 0.999 999 735 195 070 627 84 × 2 = 1 + 0.999 999 470 390 141 255 68;
  • 39) 0.999 999 470 390 141 255 68 × 2 = 1 + 0.999 998 940 780 282 511 36;
  • 40) 0.999 998 940 780 282 511 36 × 2 = 1 + 0.999 997 881 560 565 022 72;
  • 41) 0.999 997 881 560 565 022 72 × 2 = 1 + 0.999 995 763 121 130 045 44;
  • 42) 0.999 995 763 121 130 045 44 × 2 = 1 + 0.999 991 526 242 260 090 88;
  • 43) 0.999 991 526 242 260 090 88 × 2 = 1 + 0.999 983 052 484 520 181 76;
  • 44) 0.999 983 052 484 520 181 76 × 2 = 1 + 0.999 966 104 969 040 363 52;
  • 45) 0.999 966 104 969 040 363 52 × 2 = 1 + 0.999 932 209 938 080 727 04;
  • 46) 0.999 932 209 938 080 727 04 × 2 = 1 + 0.999 864 419 876 161 454 08;
  • 47) 0.999 864 419 876 161 454 08 × 2 = 1 + 0.999 728 839 752 322 908 16;
  • 48) 0.999 728 839 752 322 908 16 × 2 = 1 + 0.999 457 679 504 645 816 32;
  • 49) 0.999 457 679 504 645 816 32 × 2 = 1 + 0.998 915 359 009 291 632 64;
  • 50) 0.998 915 359 009 291 632 64 × 2 = 1 + 0.997 830 718 018 583 265 28;
  • 51) 0.997 830 718 018 583 265 28 × 2 = 1 + 0.995 661 436 037 166 530 56;
  • 52) 0.995 661 436 037 166 530 56 × 2 = 1 + 0.991 322 872 074 333 061 12;
  • 53) 0.991 322 872 074 333 061 12 × 2 = 1 + 0.982 645 744 148 666 122 24;
  • 54) 0.982 645 744 148 666 122 24 × 2 = 1 + 0.965 291 488 297 332 244 48;

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:


0.000 000 000 742 147 674 72(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 72(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:


0.000 000 000 742 147 674 72(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 -0.000 000 000 742 147 674 72 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