-0.000 000 000 829 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 829(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
-0.000 000 000 829(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:

|-0.000 000 000 829| = 0.000 000 000 829


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: 0.000 000 000 829.

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) 0.000 000 000 829 × 2 = 0 + 0.000 000 001 658;
  • 2) 0.000 000 001 658 × 2 = 0 + 0.000 000 003 316;
  • 3) 0.000 000 003 316 × 2 = 0 + 0.000 000 006 632;
  • 4) 0.000 000 006 632 × 2 = 0 + 0.000 000 013 264;
  • 5) 0.000 000 013 264 × 2 = 0 + 0.000 000 026 528;
  • 6) 0.000 000 026 528 × 2 = 0 + 0.000 000 053 056;
  • 7) 0.000 000 053 056 × 2 = 0 + 0.000 000 106 112;
  • 8) 0.000 000 106 112 × 2 = 0 + 0.000 000 212 224;
  • 9) 0.000 000 212 224 × 2 = 0 + 0.000 000 424 448;
  • 10) 0.000 000 424 448 × 2 = 0 + 0.000 000 848 896;
  • 11) 0.000 000 848 896 × 2 = 0 + 0.000 001 697 792;
  • 12) 0.000 001 697 792 × 2 = 0 + 0.000 003 395 584;
  • 13) 0.000 003 395 584 × 2 = 0 + 0.000 006 791 168;
  • 14) 0.000 006 791 168 × 2 = 0 + 0.000 013 582 336;
  • 15) 0.000 013 582 336 × 2 = 0 + 0.000 027 164 672;
  • 16) 0.000 027 164 672 × 2 = 0 + 0.000 054 329 344;
  • 17) 0.000 054 329 344 × 2 = 0 + 0.000 108 658 688;
  • 18) 0.000 108 658 688 × 2 = 0 + 0.000 217 317 376;
  • 19) 0.000 217 317 376 × 2 = 0 + 0.000 434 634 752;
  • 20) 0.000 434 634 752 × 2 = 0 + 0.000 869 269 504;
  • 21) 0.000 869 269 504 × 2 = 0 + 0.001 738 539 008;
  • 22) 0.001 738 539 008 × 2 = 0 + 0.003 477 078 016;
  • 23) 0.003 477 078 016 × 2 = 0 + 0.006 954 156 032;
  • 24) 0.006 954 156 032 × 2 = 0 + 0.013 908 312 064;
  • 25) 0.013 908 312 064 × 2 = 0 + 0.027 816 624 128;
  • 26) 0.027 816 624 128 × 2 = 0 + 0.055 633 248 256;
  • 27) 0.055 633 248 256 × 2 = 0 + 0.111 266 496 512;
  • 28) 0.111 266 496 512 × 2 = 0 + 0.222 532 993 024;
  • 29) 0.222 532 993 024 × 2 = 0 + 0.445 065 986 048;
  • 30) 0.445 065 986 048 × 2 = 0 + 0.890 131 972 096;
  • 31) 0.890 131 972 096 × 2 = 1 + 0.780 263 944 192;
  • 32) 0.780 263 944 192 × 2 = 1 + 0.560 527 888 384;
  • 33) 0.560 527 888 384 × 2 = 1 + 0.121 055 776 768;
  • 34) 0.121 055 776 768 × 2 = 0 + 0.242 111 553 536;
  • 35) 0.242 111 553 536 × 2 = 0 + 0.484 223 107 072;
  • 36) 0.484 223 107 072 × 2 = 0 + 0.968 446 214 144;
  • 37) 0.968 446 214 144 × 2 = 1 + 0.936 892 428 288;
  • 38) 0.936 892 428 288 × 2 = 1 + 0.873 784 856 576;
  • 39) 0.873 784 856 576 × 2 = 1 + 0.747 569 713 152;
  • 40) 0.747 569 713 152 × 2 = 1 + 0.495 139 426 304;
  • 41) 0.495 139 426 304 × 2 = 0 + 0.990 278 852 608;
  • 42) 0.990 278 852 608 × 2 = 1 + 0.980 557 705 216;
  • 43) 0.980 557 705 216 × 2 = 1 + 0.961 115 410 432;
  • 44) 0.961 115 410 432 × 2 = 1 + 0.922 230 820 864;
  • 45) 0.922 230 820 864 × 2 = 1 + 0.844 461 641 728;
  • 46) 0.844 461 641 728 × 2 = 1 + 0.688 923 283 456;
  • 47) 0.688 923 283 456 × 2 = 1 + 0.377 846 566 912;
  • 48) 0.377 846 566 912 × 2 = 0 + 0.755 693 133 824;
  • 49) 0.755 693 133 824 × 2 = 1 + 0.511 386 267 648;
  • 50) 0.511 386 267 648 × 2 = 1 + 0.022 772 535 296;
  • 51) 0.022 772 535 296 × 2 = 0 + 0.045 545 070 592;
  • 52) 0.045 545 070 592 × 2 = 0 + 0.091 090 141 184;
  • 53) 0.091 090 141 184 × 2 = 0 + 0.182 180 282 368;
  • 54) 0.182 180 282 368 × 2 = 0 + 0.364 360 564 736;

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:


0.000 000 000 829(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1111 0111 1110 1100 00(2)

6. Positive number before normalization:

0.000 000 000 829(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1111 0111 1110 1100 00(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:


0.000 000 000 829(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1111 0111 1110 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1111 0111 1110 1100 00(2) × 20 =

1.1100 0111 1011 1111 0110 000(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):
1.1100 0111 1011 1111 0110 000


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. 110 0011 1101 1111 1011 0000 =

110 0011 1101 1111 1011 0000


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) =
110 0011 1101 1111 1011 0000


Decimal number -0.000 000 000 829 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0011 1101 1111 1011 0000

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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