-0.000 000 000 742 147 674 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 61(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 61(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 61| = 0.000 000 000 742 147 674 61


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 61.

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 61 × 2 = 0 + 0.000 000 001 484 295 349 22;
  • 2) 0.000 000 001 484 295 349 22 × 2 = 0 + 0.000 000 002 968 590 698 44;
  • 3) 0.000 000 002 968 590 698 44 × 2 = 0 + 0.000 000 005 937 181 396 88;
  • 4) 0.000 000 005 937 181 396 88 × 2 = 0 + 0.000 000 011 874 362 793 76;
  • 5) 0.000 000 011 874 362 793 76 × 2 = 0 + 0.000 000 023 748 725 587 52;
  • 6) 0.000 000 023 748 725 587 52 × 2 = 0 + 0.000 000 047 497 451 175 04;
  • 7) 0.000 000 047 497 451 175 04 × 2 = 0 + 0.000 000 094 994 902 350 08;
  • 8) 0.000 000 094 994 902 350 08 × 2 = 0 + 0.000 000 189 989 804 700 16;
  • 9) 0.000 000 189 989 804 700 16 × 2 = 0 + 0.000 000 379 979 609 400 32;
  • 10) 0.000 000 379 979 609 400 32 × 2 = 0 + 0.000 000 759 959 218 800 64;
  • 11) 0.000 000 759 959 218 800 64 × 2 = 0 + 0.000 001 519 918 437 601 28;
  • 12) 0.000 001 519 918 437 601 28 × 2 = 0 + 0.000 003 039 836 875 202 56;
  • 13) 0.000 003 039 836 875 202 56 × 2 = 0 + 0.000 006 079 673 750 405 12;
  • 14) 0.000 006 079 673 750 405 12 × 2 = 0 + 0.000 012 159 347 500 810 24;
  • 15) 0.000 012 159 347 500 810 24 × 2 = 0 + 0.000 024 318 695 001 620 48;
  • 16) 0.000 024 318 695 001 620 48 × 2 = 0 + 0.000 048 637 390 003 240 96;
  • 17) 0.000 048 637 390 003 240 96 × 2 = 0 + 0.000 097 274 780 006 481 92;
  • 18) 0.000 097 274 780 006 481 92 × 2 = 0 + 0.000 194 549 560 012 963 84;
  • 19) 0.000 194 549 560 012 963 84 × 2 = 0 + 0.000 389 099 120 025 927 68;
  • 20) 0.000 389 099 120 025 927 68 × 2 = 0 + 0.000 778 198 240 051 855 36;
  • 21) 0.000 778 198 240 051 855 36 × 2 = 0 + 0.001 556 396 480 103 710 72;
  • 22) 0.001 556 396 480 103 710 72 × 2 = 0 + 0.003 112 792 960 207 421 44;
  • 23) 0.003 112 792 960 207 421 44 × 2 = 0 + 0.006 225 585 920 414 842 88;
  • 24) 0.006 225 585 920 414 842 88 × 2 = 0 + 0.012 451 171 840 829 685 76;
  • 25) 0.012 451 171 840 829 685 76 × 2 = 0 + 0.024 902 343 681 659 371 52;
  • 26) 0.024 902 343 681 659 371 52 × 2 = 0 + 0.049 804 687 363 318 743 04;
  • 27) 0.049 804 687 363 318 743 04 × 2 = 0 + 0.099 609 374 726 637 486 08;
  • 28) 0.099 609 374 726 637 486 08 × 2 = 0 + 0.199 218 749 453 274 972 16;
  • 29) 0.199 218 749 453 274 972 16 × 2 = 0 + 0.398 437 498 906 549 944 32;
  • 30) 0.398 437 498 906 549 944 32 × 2 = 0 + 0.796 874 997 813 099 888 64;
  • 31) 0.796 874 997 813 099 888 64 × 2 = 1 + 0.593 749 995 626 199 777 28;
  • 32) 0.593 749 995 626 199 777 28 × 2 = 1 + 0.187 499 991 252 399 554 56;
  • 33) 0.187 499 991 252 399 554 56 × 2 = 0 + 0.374 999 982 504 799 109 12;
  • 34) 0.374 999 982 504 799 109 12 × 2 = 0 + 0.749 999 965 009 598 218 24;
  • 35) 0.749 999 965 009 598 218 24 × 2 = 1 + 0.499 999 930 019 196 436 48;
  • 36) 0.499 999 930 019 196 436 48 × 2 = 0 + 0.999 999 860 038 392 872 96;
  • 37) 0.999 999 860 038 392 872 96 × 2 = 1 + 0.999 999 720 076 785 745 92;
  • 38) 0.999 999 720 076 785 745 92 × 2 = 1 + 0.999 999 440 153 571 491 84;
  • 39) 0.999 999 440 153 571 491 84 × 2 = 1 + 0.999 998 880 307 142 983 68;
  • 40) 0.999 998 880 307 142 983 68 × 2 = 1 + 0.999 997 760 614 285 967 36;
  • 41) 0.999 997 760 614 285 967 36 × 2 = 1 + 0.999 995 521 228 571 934 72;
  • 42) 0.999 995 521 228 571 934 72 × 2 = 1 + 0.999 991 042 457 143 869 44;
  • 43) 0.999 991 042 457 143 869 44 × 2 = 1 + 0.999 982 084 914 287 738 88;
  • 44) 0.999 982 084 914 287 738 88 × 2 = 1 + 0.999 964 169 828 575 477 76;
  • 45) 0.999 964 169 828 575 477 76 × 2 = 1 + 0.999 928 339 657 150 955 52;
  • 46) 0.999 928 339 657 150 955 52 × 2 = 1 + 0.999 856 679 314 301 911 04;
  • 47) 0.999 856 679 314 301 911 04 × 2 = 1 + 0.999 713 358 628 603 822 08;
  • 48) 0.999 713 358 628 603 822 08 × 2 = 1 + 0.999 426 717 257 207 644 16;
  • 49) 0.999 426 717 257 207 644 16 × 2 = 1 + 0.998 853 434 514 415 288 32;
  • 50) 0.998 853 434 514 415 288 32 × 2 = 1 + 0.997 706 869 028 830 576 64;
  • 51) 0.997 706 869 028 830 576 64 × 2 = 1 + 0.995 413 738 057 661 153 28;
  • 52) 0.995 413 738 057 661 153 28 × 2 = 1 + 0.990 827 476 115 322 306 56;
  • 53) 0.990 827 476 115 322 306 56 × 2 = 1 + 0.981 654 952 230 644 613 12;
  • 54) 0.981 654 952 230 644 613 12 × 2 = 1 + 0.963 309 904 461 289 226 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:


0.000 000 000 742 147 674 61(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 61(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 61(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 61 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