-0.000 000 000 742 147 674 44 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 44(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 44(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 44| = 0.000 000 000 742 147 674 44


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 44.

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 44 × 2 = 0 + 0.000 000 001 484 295 348 88;
  • 2) 0.000 000 001 484 295 348 88 × 2 = 0 + 0.000 000 002 968 590 697 76;
  • 3) 0.000 000 002 968 590 697 76 × 2 = 0 + 0.000 000 005 937 181 395 52;
  • 4) 0.000 000 005 937 181 395 52 × 2 = 0 + 0.000 000 011 874 362 791 04;
  • 5) 0.000 000 011 874 362 791 04 × 2 = 0 + 0.000 000 023 748 725 582 08;
  • 6) 0.000 000 023 748 725 582 08 × 2 = 0 + 0.000 000 047 497 451 164 16;
  • 7) 0.000 000 047 497 451 164 16 × 2 = 0 + 0.000 000 094 994 902 328 32;
  • 8) 0.000 000 094 994 902 328 32 × 2 = 0 + 0.000 000 189 989 804 656 64;
  • 9) 0.000 000 189 989 804 656 64 × 2 = 0 + 0.000 000 379 979 609 313 28;
  • 10) 0.000 000 379 979 609 313 28 × 2 = 0 + 0.000 000 759 959 218 626 56;
  • 11) 0.000 000 759 959 218 626 56 × 2 = 0 + 0.000 001 519 918 437 253 12;
  • 12) 0.000 001 519 918 437 253 12 × 2 = 0 + 0.000 003 039 836 874 506 24;
  • 13) 0.000 003 039 836 874 506 24 × 2 = 0 + 0.000 006 079 673 749 012 48;
  • 14) 0.000 006 079 673 749 012 48 × 2 = 0 + 0.000 012 159 347 498 024 96;
  • 15) 0.000 012 159 347 498 024 96 × 2 = 0 + 0.000 024 318 694 996 049 92;
  • 16) 0.000 024 318 694 996 049 92 × 2 = 0 + 0.000 048 637 389 992 099 84;
  • 17) 0.000 048 637 389 992 099 84 × 2 = 0 + 0.000 097 274 779 984 199 68;
  • 18) 0.000 097 274 779 984 199 68 × 2 = 0 + 0.000 194 549 559 968 399 36;
  • 19) 0.000 194 549 559 968 399 36 × 2 = 0 + 0.000 389 099 119 936 798 72;
  • 20) 0.000 389 099 119 936 798 72 × 2 = 0 + 0.000 778 198 239 873 597 44;
  • 21) 0.000 778 198 239 873 597 44 × 2 = 0 + 0.001 556 396 479 747 194 88;
  • 22) 0.001 556 396 479 747 194 88 × 2 = 0 + 0.003 112 792 959 494 389 76;
  • 23) 0.003 112 792 959 494 389 76 × 2 = 0 + 0.006 225 585 918 988 779 52;
  • 24) 0.006 225 585 918 988 779 52 × 2 = 0 + 0.012 451 171 837 977 559 04;
  • 25) 0.012 451 171 837 977 559 04 × 2 = 0 + 0.024 902 343 675 955 118 08;
  • 26) 0.024 902 343 675 955 118 08 × 2 = 0 + 0.049 804 687 351 910 236 16;
  • 27) 0.049 804 687 351 910 236 16 × 2 = 0 + 0.099 609 374 703 820 472 32;
  • 28) 0.099 609 374 703 820 472 32 × 2 = 0 + 0.199 218 749 407 640 944 64;
  • 29) 0.199 218 749 407 640 944 64 × 2 = 0 + 0.398 437 498 815 281 889 28;
  • 30) 0.398 437 498 815 281 889 28 × 2 = 0 + 0.796 874 997 630 563 778 56;
  • 31) 0.796 874 997 630 563 778 56 × 2 = 1 + 0.593 749 995 261 127 557 12;
  • 32) 0.593 749 995 261 127 557 12 × 2 = 1 + 0.187 499 990 522 255 114 24;
  • 33) 0.187 499 990 522 255 114 24 × 2 = 0 + 0.374 999 981 044 510 228 48;
  • 34) 0.374 999 981 044 510 228 48 × 2 = 0 + 0.749 999 962 089 020 456 96;
  • 35) 0.749 999 962 089 020 456 96 × 2 = 1 + 0.499 999 924 178 040 913 92;
  • 36) 0.499 999 924 178 040 913 92 × 2 = 0 + 0.999 999 848 356 081 827 84;
  • 37) 0.999 999 848 356 081 827 84 × 2 = 1 + 0.999 999 696 712 163 655 68;
  • 38) 0.999 999 696 712 163 655 68 × 2 = 1 + 0.999 999 393 424 327 311 36;
  • 39) 0.999 999 393 424 327 311 36 × 2 = 1 + 0.999 998 786 848 654 622 72;
  • 40) 0.999 998 786 848 654 622 72 × 2 = 1 + 0.999 997 573 697 309 245 44;
  • 41) 0.999 997 573 697 309 245 44 × 2 = 1 + 0.999 995 147 394 618 490 88;
  • 42) 0.999 995 147 394 618 490 88 × 2 = 1 + 0.999 990 294 789 236 981 76;
  • 43) 0.999 990 294 789 236 981 76 × 2 = 1 + 0.999 980 589 578 473 963 52;
  • 44) 0.999 980 589 578 473 963 52 × 2 = 1 + 0.999 961 179 156 947 927 04;
  • 45) 0.999 961 179 156 947 927 04 × 2 = 1 + 0.999 922 358 313 895 854 08;
  • 46) 0.999 922 358 313 895 854 08 × 2 = 1 + 0.999 844 716 627 791 708 16;
  • 47) 0.999 844 716 627 791 708 16 × 2 = 1 + 0.999 689 433 255 583 416 32;
  • 48) 0.999 689 433 255 583 416 32 × 2 = 1 + 0.999 378 866 511 166 832 64;
  • 49) 0.999 378 866 511 166 832 64 × 2 = 1 + 0.998 757 733 022 333 665 28;
  • 50) 0.998 757 733 022 333 665 28 × 2 = 1 + 0.997 515 466 044 667 330 56;
  • 51) 0.997 515 466 044 667 330 56 × 2 = 1 + 0.995 030 932 089 334 661 12;
  • 52) 0.995 030 932 089 334 661 12 × 2 = 1 + 0.990 061 864 178 669 322 24;
  • 53) 0.990 061 864 178 669 322 24 × 2 = 1 + 0.980 123 728 357 338 644 48;
  • 54) 0.980 123 728 357 338 644 48 × 2 = 1 + 0.960 247 456 714 677 288 96;

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 44(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 44(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 44(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 44 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