-0.000 000 000 742 147 674 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 5(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 5(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 5| = 0.000 000 000 742 147 674 5


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 5.

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 5 × 2 = 0 + 0.000 000 001 484 295 349;
  • 2) 0.000 000 001 484 295 349 × 2 = 0 + 0.000 000 002 968 590 698;
  • 3) 0.000 000 002 968 590 698 × 2 = 0 + 0.000 000 005 937 181 396;
  • 4) 0.000 000 005 937 181 396 × 2 = 0 + 0.000 000 011 874 362 792;
  • 5) 0.000 000 011 874 362 792 × 2 = 0 + 0.000 000 023 748 725 584;
  • 6) 0.000 000 023 748 725 584 × 2 = 0 + 0.000 000 047 497 451 168;
  • 7) 0.000 000 047 497 451 168 × 2 = 0 + 0.000 000 094 994 902 336;
  • 8) 0.000 000 094 994 902 336 × 2 = 0 + 0.000 000 189 989 804 672;
  • 9) 0.000 000 189 989 804 672 × 2 = 0 + 0.000 000 379 979 609 344;
  • 10) 0.000 000 379 979 609 344 × 2 = 0 + 0.000 000 759 959 218 688;
  • 11) 0.000 000 759 959 218 688 × 2 = 0 + 0.000 001 519 918 437 376;
  • 12) 0.000 001 519 918 437 376 × 2 = 0 + 0.000 003 039 836 874 752;
  • 13) 0.000 003 039 836 874 752 × 2 = 0 + 0.000 006 079 673 749 504;
  • 14) 0.000 006 079 673 749 504 × 2 = 0 + 0.000 012 159 347 499 008;
  • 15) 0.000 012 159 347 499 008 × 2 = 0 + 0.000 024 318 694 998 016;
  • 16) 0.000 024 318 694 998 016 × 2 = 0 + 0.000 048 637 389 996 032;
  • 17) 0.000 048 637 389 996 032 × 2 = 0 + 0.000 097 274 779 992 064;
  • 18) 0.000 097 274 779 992 064 × 2 = 0 + 0.000 194 549 559 984 128;
  • 19) 0.000 194 549 559 984 128 × 2 = 0 + 0.000 389 099 119 968 256;
  • 20) 0.000 389 099 119 968 256 × 2 = 0 + 0.000 778 198 239 936 512;
  • 21) 0.000 778 198 239 936 512 × 2 = 0 + 0.001 556 396 479 873 024;
  • 22) 0.001 556 396 479 873 024 × 2 = 0 + 0.003 112 792 959 746 048;
  • 23) 0.003 112 792 959 746 048 × 2 = 0 + 0.006 225 585 919 492 096;
  • 24) 0.006 225 585 919 492 096 × 2 = 0 + 0.012 451 171 838 984 192;
  • 25) 0.012 451 171 838 984 192 × 2 = 0 + 0.024 902 343 677 968 384;
  • 26) 0.024 902 343 677 968 384 × 2 = 0 + 0.049 804 687 355 936 768;
  • 27) 0.049 804 687 355 936 768 × 2 = 0 + 0.099 609 374 711 873 536;
  • 28) 0.099 609 374 711 873 536 × 2 = 0 + 0.199 218 749 423 747 072;
  • 29) 0.199 218 749 423 747 072 × 2 = 0 + 0.398 437 498 847 494 144;
  • 30) 0.398 437 498 847 494 144 × 2 = 0 + 0.796 874 997 694 988 288;
  • 31) 0.796 874 997 694 988 288 × 2 = 1 + 0.593 749 995 389 976 576;
  • 32) 0.593 749 995 389 976 576 × 2 = 1 + 0.187 499 990 779 953 152;
  • 33) 0.187 499 990 779 953 152 × 2 = 0 + 0.374 999 981 559 906 304;
  • 34) 0.374 999 981 559 906 304 × 2 = 0 + 0.749 999 963 119 812 608;
  • 35) 0.749 999 963 119 812 608 × 2 = 1 + 0.499 999 926 239 625 216;
  • 36) 0.499 999 926 239 625 216 × 2 = 0 + 0.999 999 852 479 250 432;
  • 37) 0.999 999 852 479 250 432 × 2 = 1 + 0.999 999 704 958 500 864;
  • 38) 0.999 999 704 958 500 864 × 2 = 1 + 0.999 999 409 917 001 728;
  • 39) 0.999 999 409 917 001 728 × 2 = 1 + 0.999 998 819 834 003 456;
  • 40) 0.999 998 819 834 003 456 × 2 = 1 + 0.999 997 639 668 006 912;
  • 41) 0.999 997 639 668 006 912 × 2 = 1 + 0.999 995 279 336 013 824;
  • 42) 0.999 995 279 336 013 824 × 2 = 1 + 0.999 990 558 672 027 648;
  • 43) 0.999 990 558 672 027 648 × 2 = 1 + 0.999 981 117 344 055 296;
  • 44) 0.999 981 117 344 055 296 × 2 = 1 + 0.999 962 234 688 110 592;
  • 45) 0.999 962 234 688 110 592 × 2 = 1 + 0.999 924 469 376 221 184;
  • 46) 0.999 924 469 376 221 184 × 2 = 1 + 0.999 848 938 752 442 368;
  • 47) 0.999 848 938 752 442 368 × 2 = 1 + 0.999 697 877 504 884 736;
  • 48) 0.999 697 877 504 884 736 × 2 = 1 + 0.999 395 755 009 769 472;
  • 49) 0.999 395 755 009 769 472 × 2 = 1 + 0.998 791 510 019 538 944;
  • 50) 0.998 791 510 019 538 944 × 2 = 1 + 0.997 583 020 039 077 888;
  • 51) 0.997 583 020 039 077 888 × 2 = 1 + 0.995 166 040 078 155 776;
  • 52) 0.995 166 040 078 155 776 × 2 = 1 + 0.990 332 080 156 311 552;
  • 53) 0.990 332 080 156 311 552 × 2 = 1 + 0.980 664 160 312 623 104;
  • 54) 0.980 664 160 312 623 104 × 2 = 1 + 0.961 328 320 625 246 208;

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 5(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 5(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 5(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 5 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