-0.000 000 000 742 149 17 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 17(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
-0.000 000 000 742 149 17(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:

|-0.000 000 000 742 149 17| = 0.000 000 000 742 149 17


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: 0.000 000 000 742 149 17.

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) 0.000 000 000 742 149 17 × 2 = 0 + 0.000 000 001 484 298 34;
  • 2) 0.000 000 001 484 298 34 × 2 = 0 + 0.000 000 002 968 596 68;
  • 3) 0.000 000 002 968 596 68 × 2 = 0 + 0.000 000 005 937 193 36;
  • 4) 0.000 000 005 937 193 36 × 2 = 0 + 0.000 000 011 874 386 72;
  • 5) 0.000 000 011 874 386 72 × 2 = 0 + 0.000 000 023 748 773 44;
  • 6) 0.000 000 023 748 773 44 × 2 = 0 + 0.000 000 047 497 546 88;
  • 7) 0.000 000 047 497 546 88 × 2 = 0 + 0.000 000 094 995 093 76;
  • 8) 0.000 000 094 995 093 76 × 2 = 0 + 0.000 000 189 990 187 52;
  • 9) 0.000 000 189 990 187 52 × 2 = 0 + 0.000 000 379 980 375 04;
  • 10) 0.000 000 379 980 375 04 × 2 = 0 + 0.000 000 759 960 750 08;
  • 11) 0.000 000 759 960 750 08 × 2 = 0 + 0.000 001 519 921 500 16;
  • 12) 0.000 001 519 921 500 16 × 2 = 0 + 0.000 003 039 843 000 32;
  • 13) 0.000 003 039 843 000 32 × 2 = 0 + 0.000 006 079 686 000 64;
  • 14) 0.000 006 079 686 000 64 × 2 = 0 + 0.000 012 159 372 001 28;
  • 15) 0.000 012 159 372 001 28 × 2 = 0 + 0.000 024 318 744 002 56;
  • 16) 0.000 024 318 744 002 56 × 2 = 0 + 0.000 048 637 488 005 12;
  • 17) 0.000 048 637 488 005 12 × 2 = 0 + 0.000 097 274 976 010 24;
  • 18) 0.000 097 274 976 010 24 × 2 = 0 + 0.000 194 549 952 020 48;
  • 19) 0.000 194 549 952 020 48 × 2 = 0 + 0.000 389 099 904 040 96;
  • 20) 0.000 389 099 904 040 96 × 2 = 0 + 0.000 778 199 808 081 92;
  • 21) 0.000 778 199 808 081 92 × 2 = 0 + 0.001 556 399 616 163 84;
  • 22) 0.001 556 399 616 163 84 × 2 = 0 + 0.003 112 799 232 327 68;
  • 23) 0.003 112 799 232 327 68 × 2 = 0 + 0.006 225 598 464 655 36;
  • 24) 0.006 225 598 464 655 36 × 2 = 0 + 0.012 451 196 929 310 72;
  • 25) 0.012 451 196 929 310 72 × 2 = 0 + 0.024 902 393 858 621 44;
  • 26) 0.024 902 393 858 621 44 × 2 = 0 + 0.049 804 787 717 242 88;
  • 27) 0.049 804 787 717 242 88 × 2 = 0 + 0.099 609 575 434 485 76;
  • 28) 0.099 609 575 434 485 76 × 2 = 0 + 0.199 219 150 868 971 52;
  • 29) 0.199 219 150 868 971 52 × 2 = 0 + 0.398 438 301 737 943 04;
  • 30) 0.398 438 301 737 943 04 × 2 = 0 + 0.796 876 603 475 886 08;
  • 31) 0.796 876 603 475 886 08 × 2 = 1 + 0.593 753 206 951 772 16;
  • 32) 0.593 753 206 951 772 16 × 2 = 1 + 0.187 506 413 903 544 32;
  • 33) 0.187 506 413 903 544 32 × 2 = 0 + 0.375 012 827 807 088 64;
  • 34) 0.375 012 827 807 088 64 × 2 = 0 + 0.750 025 655 614 177 28;
  • 35) 0.750 025 655 614 177 28 × 2 = 1 + 0.500 051 311 228 354 56;
  • 36) 0.500 051 311 228 354 56 × 2 = 1 + 0.000 102 622 456 709 12;
  • 37) 0.000 102 622 456 709 12 × 2 = 0 + 0.000 205 244 913 418 24;
  • 38) 0.000 205 244 913 418 24 × 2 = 0 + 0.000 410 489 826 836 48;
  • 39) 0.000 410 489 826 836 48 × 2 = 0 + 0.000 820 979 653 672 96;
  • 40) 0.000 820 979 653 672 96 × 2 = 0 + 0.001 641 959 307 345 92;
  • 41) 0.001 641 959 307 345 92 × 2 = 0 + 0.003 283 918 614 691 84;
  • 42) 0.003 283 918 614 691 84 × 2 = 0 + 0.006 567 837 229 383 68;
  • 43) 0.006 567 837 229 383 68 × 2 = 0 + 0.013 135 674 458 767 36;
  • 44) 0.013 135 674 458 767 36 × 2 = 0 + 0.026 271 348 917 534 72;
  • 45) 0.026 271 348 917 534 72 × 2 = 0 + 0.052 542 697 835 069 44;
  • 46) 0.052 542 697 835 069 44 × 2 = 0 + 0.105 085 395 670 138 88;
  • 47) 0.105 085 395 670 138 88 × 2 = 0 + 0.210 170 791 340 277 76;
  • 48) 0.210 170 791 340 277 76 × 2 = 0 + 0.420 341 582 680 555 52;
  • 49) 0.420 341 582 680 555 52 × 2 = 0 + 0.840 683 165 361 111 04;
  • 50) 0.840 683 165 361 111 04 × 2 = 1 + 0.681 366 330 722 222 08;
  • 51) 0.681 366 330 722 222 08 × 2 = 1 + 0.362 732 661 444 444 16;
  • 52) 0.362 732 661 444 444 16 × 2 = 0 + 0.725 465 322 888 888 32;
  • 53) 0.725 465 322 888 888 32 × 2 = 1 + 0.450 930 645 777 776 64;
  • 54) 0.450 930 645 777 776 64 × 2 = 0 + 0.901 861 291 555 553 28;

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:


0.000 000 000 742 149 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2)

6. Positive number before normalization:

0.000 000 000 742 149 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(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:


0.000 000 000 742 149 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 10(2) × 20 =

1.1001 1000 0000 0000 0011 010(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 1000 0000 0000 0011 010


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 1100 0000 0000 0001 1010 =

100 1100 0000 0000 0001 1010


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 1100 0000 0000 0001 1010


Decimal number -0.000 000 000 742 149 17 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 1010

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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