-0.000 000 000 742 147 673 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 66(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
-0.000 000 000 742 147 673 66(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:

|-0.000 000 000 742 147 673 66| = 0.000 000 000 742 147 673 66


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: 0.000 000 000 742 147 673 66.

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) 0.000 000 000 742 147 673 66 × 2 = 0 + 0.000 000 001 484 295 347 32;
  • 2) 0.000 000 001 484 295 347 32 × 2 = 0 + 0.000 000 002 968 590 694 64;
  • 3) 0.000 000 002 968 590 694 64 × 2 = 0 + 0.000 000 005 937 181 389 28;
  • 4) 0.000 000 005 937 181 389 28 × 2 = 0 + 0.000 000 011 874 362 778 56;
  • 5) 0.000 000 011 874 362 778 56 × 2 = 0 + 0.000 000 023 748 725 557 12;
  • 6) 0.000 000 023 748 725 557 12 × 2 = 0 + 0.000 000 047 497 451 114 24;
  • 7) 0.000 000 047 497 451 114 24 × 2 = 0 + 0.000 000 094 994 902 228 48;
  • 8) 0.000 000 094 994 902 228 48 × 2 = 0 + 0.000 000 189 989 804 456 96;
  • 9) 0.000 000 189 989 804 456 96 × 2 = 0 + 0.000 000 379 979 608 913 92;
  • 10) 0.000 000 379 979 608 913 92 × 2 = 0 + 0.000 000 759 959 217 827 84;
  • 11) 0.000 000 759 959 217 827 84 × 2 = 0 + 0.000 001 519 918 435 655 68;
  • 12) 0.000 001 519 918 435 655 68 × 2 = 0 + 0.000 003 039 836 871 311 36;
  • 13) 0.000 003 039 836 871 311 36 × 2 = 0 + 0.000 006 079 673 742 622 72;
  • 14) 0.000 006 079 673 742 622 72 × 2 = 0 + 0.000 012 159 347 485 245 44;
  • 15) 0.000 012 159 347 485 245 44 × 2 = 0 + 0.000 024 318 694 970 490 88;
  • 16) 0.000 024 318 694 970 490 88 × 2 = 0 + 0.000 048 637 389 940 981 76;
  • 17) 0.000 048 637 389 940 981 76 × 2 = 0 + 0.000 097 274 779 881 963 52;
  • 18) 0.000 097 274 779 881 963 52 × 2 = 0 + 0.000 194 549 559 763 927 04;
  • 19) 0.000 194 549 559 763 927 04 × 2 = 0 + 0.000 389 099 119 527 854 08;
  • 20) 0.000 389 099 119 527 854 08 × 2 = 0 + 0.000 778 198 239 055 708 16;
  • 21) 0.000 778 198 239 055 708 16 × 2 = 0 + 0.001 556 396 478 111 416 32;
  • 22) 0.001 556 396 478 111 416 32 × 2 = 0 + 0.003 112 792 956 222 832 64;
  • 23) 0.003 112 792 956 222 832 64 × 2 = 0 + 0.006 225 585 912 445 665 28;
  • 24) 0.006 225 585 912 445 665 28 × 2 = 0 + 0.012 451 171 824 891 330 56;
  • 25) 0.012 451 171 824 891 330 56 × 2 = 0 + 0.024 902 343 649 782 661 12;
  • 26) 0.024 902 343 649 782 661 12 × 2 = 0 + 0.049 804 687 299 565 322 24;
  • 27) 0.049 804 687 299 565 322 24 × 2 = 0 + 0.099 609 374 599 130 644 48;
  • 28) 0.099 609 374 599 130 644 48 × 2 = 0 + 0.199 218 749 198 261 288 96;
  • 29) 0.199 218 749 198 261 288 96 × 2 = 0 + 0.398 437 498 396 522 577 92;
  • 30) 0.398 437 498 396 522 577 92 × 2 = 0 + 0.796 874 996 793 045 155 84;
  • 31) 0.796 874 996 793 045 155 84 × 2 = 1 + 0.593 749 993 586 090 311 68;
  • 32) 0.593 749 993 586 090 311 68 × 2 = 1 + 0.187 499 987 172 180 623 36;
  • 33) 0.187 499 987 172 180 623 36 × 2 = 0 + 0.374 999 974 344 361 246 72;
  • 34) 0.374 999 974 344 361 246 72 × 2 = 0 + 0.749 999 948 688 722 493 44;
  • 35) 0.749 999 948 688 722 493 44 × 2 = 1 + 0.499 999 897 377 444 986 88;
  • 36) 0.499 999 897 377 444 986 88 × 2 = 0 + 0.999 999 794 754 889 973 76;
  • 37) 0.999 999 794 754 889 973 76 × 2 = 1 + 0.999 999 589 509 779 947 52;
  • 38) 0.999 999 589 509 779 947 52 × 2 = 1 + 0.999 999 179 019 559 895 04;
  • 39) 0.999 999 179 019 559 895 04 × 2 = 1 + 0.999 998 358 039 119 790 08;
  • 40) 0.999 998 358 039 119 790 08 × 2 = 1 + 0.999 996 716 078 239 580 16;
  • 41) 0.999 996 716 078 239 580 16 × 2 = 1 + 0.999 993 432 156 479 160 32;
  • 42) 0.999 993 432 156 479 160 32 × 2 = 1 + 0.999 986 864 312 958 320 64;
  • 43) 0.999 986 864 312 958 320 64 × 2 = 1 + 0.999 973 728 625 916 641 28;
  • 44) 0.999 973 728 625 916 641 28 × 2 = 1 + 0.999 947 457 251 833 282 56;
  • 45) 0.999 947 457 251 833 282 56 × 2 = 1 + 0.999 894 914 503 666 565 12;
  • 46) 0.999 894 914 503 666 565 12 × 2 = 1 + 0.999 789 829 007 333 130 24;
  • 47) 0.999 789 829 007 333 130 24 × 2 = 1 + 0.999 579 658 014 666 260 48;
  • 48) 0.999 579 658 014 666 260 48 × 2 = 1 + 0.999 159 316 029 332 520 96;
  • 49) 0.999 159 316 029 332 520 96 × 2 = 1 + 0.998 318 632 058 665 041 92;
  • 50) 0.998 318 632 058 665 041 92 × 2 = 1 + 0.996 637 264 117 330 083 84;
  • 51) 0.996 637 264 117 330 083 84 × 2 = 1 + 0.993 274 528 234 660 167 68;
  • 52) 0.993 274 528 234 660 167 68 × 2 = 1 + 0.986 549 056 469 320 335 36;
  • 53) 0.986 549 056 469 320 335 36 × 2 = 1 + 0.973 098 112 938 640 670 72;
  • 54) 0.973 098 112 938 640 670 72 × 2 = 1 + 0.946 196 225 877 281 341 44;

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:


0.000 000 000 742 147 673 66(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 66(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:


0.000 000 000 742 147 673 66(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 -0.000 000 000 742 147 673 66 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