-0.000 000 000 742 147 672 93 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 93(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
-0.000 000 000 742 147 672 93(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:

|-0.000 000 000 742 147 672 93| = 0.000 000 000 742 147 672 93


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: 0.000 000 000 742 147 672 93.

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) 0.000 000 000 742 147 672 93 × 2 = 0 + 0.000 000 001 484 295 345 86;
  • 2) 0.000 000 001 484 295 345 86 × 2 = 0 + 0.000 000 002 968 590 691 72;
  • 3) 0.000 000 002 968 590 691 72 × 2 = 0 + 0.000 000 005 937 181 383 44;
  • 4) 0.000 000 005 937 181 383 44 × 2 = 0 + 0.000 000 011 874 362 766 88;
  • 5) 0.000 000 011 874 362 766 88 × 2 = 0 + 0.000 000 023 748 725 533 76;
  • 6) 0.000 000 023 748 725 533 76 × 2 = 0 + 0.000 000 047 497 451 067 52;
  • 7) 0.000 000 047 497 451 067 52 × 2 = 0 + 0.000 000 094 994 902 135 04;
  • 8) 0.000 000 094 994 902 135 04 × 2 = 0 + 0.000 000 189 989 804 270 08;
  • 9) 0.000 000 189 989 804 270 08 × 2 = 0 + 0.000 000 379 979 608 540 16;
  • 10) 0.000 000 379 979 608 540 16 × 2 = 0 + 0.000 000 759 959 217 080 32;
  • 11) 0.000 000 759 959 217 080 32 × 2 = 0 + 0.000 001 519 918 434 160 64;
  • 12) 0.000 001 519 918 434 160 64 × 2 = 0 + 0.000 003 039 836 868 321 28;
  • 13) 0.000 003 039 836 868 321 28 × 2 = 0 + 0.000 006 079 673 736 642 56;
  • 14) 0.000 006 079 673 736 642 56 × 2 = 0 + 0.000 012 159 347 473 285 12;
  • 15) 0.000 012 159 347 473 285 12 × 2 = 0 + 0.000 024 318 694 946 570 24;
  • 16) 0.000 024 318 694 946 570 24 × 2 = 0 + 0.000 048 637 389 893 140 48;
  • 17) 0.000 048 637 389 893 140 48 × 2 = 0 + 0.000 097 274 779 786 280 96;
  • 18) 0.000 097 274 779 786 280 96 × 2 = 0 + 0.000 194 549 559 572 561 92;
  • 19) 0.000 194 549 559 572 561 92 × 2 = 0 + 0.000 389 099 119 145 123 84;
  • 20) 0.000 389 099 119 145 123 84 × 2 = 0 + 0.000 778 198 238 290 247 68;
  • 21) 0.000 778 198 238 290 247 68 × 2 = 0 + 0.001 556 396 476 580 495 36;
  • 22) 0.001 556 396 476 580 495 36 × 2 = 0 + 0.003 112 792 953 160 990 72;
  • 23) 0.003 112 792 953 160 990 72 × 2 = 0 + 0.006 225 585 906 321 981 44;
  • 24) 0.006 225 585 906 321 981 44 × 2 = 0 + 0.012 451 171 812 643 962 88;
  • 25) 0.012 451 171 812 643 962 88 × 2 = 0 + 0.024 902 343 625 287 925 76;
  • 26) 0.024 902 343 625 287 925 76 × 2 = 0 + 0.049 804 687 250 575 851 52;
  • 27) 0.049 804 687 250 575 851 52 × 2 = 0 + 0.099 609 374 501 151 703 04;
  • 28) 0.099 609 374 501 151 703 04 × 2 = 0 + 0.199 218 749 002 303 406 08;
  • 29) 0.199 218 749 002 303 406 08 × 2 = 0 + 0.398 437 498 004 606 812 16;
  • 30) 0.398 437 498 004 606 812 16 × 2 = 0 + 0.796 874 996 009 213 624 32;
  • 31) 0.796 874 996 009 213 624 32 × 2 = 1 + 0.593 749 992 018 427 248 64;
  • 32) 0.593 749 992 018 427 248 64 × 2 = 1 + 0.187 499 984 036 854 497 28;
  • 33) 0.187 499 984 036 854 497 28 × 2 = 0 + 0.374 999 968 073 708 994 56;
  • 34) 0.374 999 968 073 708 994 56 × 2 = 0 + 0.749 999 936 147 417 989 12;
  • 35) 0.749 999 936 147 417 989 12 × 2 = 1 + 0.499 999 872 294 835 978 24;
  • 36) 0.499 999 872 294 835 978 24 × 2 = 0 + 0.999 999 744 589 671 956 48;
  • 37) 0.999 999 744 589 671 956 48 × 2 = 1 + 0.999 999 489 179 343 912 96;
  • 38) 0.999 999 489 179 343 912 96 × 2 = 1 + 0.999 998 978 358 687 825 92;
  • 39) 0.999 998 978 358 687 825 92 × 2 = 1 + 0.999 997 956 717 375 651 84;
  • 40) 0.999 997 956 717 375 651 84 × 2 = 1 + 0.999 995 913 434 751 303 68;
  • 41) 0.999 995 913 434 751 303 68 × 2 = 1 + 0.999 991 826 869 502 607 36;
  • 42) 0.999 991 826 869 502 607 36 × 2 = 1 + 0.999 983 653 739 005 214 72;
  • 43) 0.999 983 653 739 005 214 72 × 2 = 1 + 0.999 967 307 478 010 429 44;
  • 44) 0.999 967 307 478 010 429 44 × 2 = 1 + 0.999 934 614 956 020 858 88;
  • 45) 0.999 934 614 956 020 858 88 × 2 = 1 + 0.999 869 229 912 041 717 76;
  • 46) 0.999 869 229 912 041 717 76 × 2 = 1 + 0.999 738 459 824 083 435 52;
  • 47) 0.999 738 459 824 083 435 52 × 2 = 1 + 0.999 476 919 648 166 871 04;
  • 48) 0.999 476 919 648 166 871 04 × 2 = 1 + 0.998 953 839 296 333 742 08;
  • 49) 0.998 953 839 296 333 742 08 × 2 = 1 + 0.997 907 678 592 667 484 16;
  • 50) 0.997 907 678 592 667 484 16 × 2 = 1 + 0.995 815 357 185 334 968 32;
  • 51) 0.995 815 357 185 334 968 32 × 2 = 1 + 0.991 630 714 370 669 936 64;
  • 52) 0.991 630 714 370 669 936 64 × 2 = 1 + 0.983 261 428 741 339 873 28;
  • 53) 0.983 261 428 741 339 873 28 × 2 = 1 + 0.966 522 857 482 679 746 56;
  • 54) 0.966 522 857 482 679 746 56 × 2 = 1 + 0.933 045 714 965 359 493 12;

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:


0.000 000 000 742 147 672 93(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 672 93(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:


0.000 000 000 742 147 672 93(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 -0.000 000 000 742 147 672 93 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