-0.000 110 011 001 100 110 011 004 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 110 011 001 100 110 011 004(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
-0.000 110 011 001 100 110 011 004(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:

|-0.000 110 011 001 100 110 011 004| = 0.000 110 011 001 100 110 011 004


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: 0.000 110 011 001 100 110 011 004.

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) 0.000 110 011 001 100 110 011 004 × 2 = 0 + 0.000 220 022 002 200 220 022 008;
  • 2) 0.000 220 022 002 200 220 022 008 × 2 = 0 + 0.000 440 044 004 400 440 044 016;
  • 3) 0.000 440 044 004 400 440 044 016 × 2 = 0 + 0.000 880 088 008 800 880 088 032;
  • 4) 0.000 880 088 008 800 880 088 032 × 2 = 0 + 0.001 760 176 017 601 760 176 064;
  • 5) 0.001 760 176 017 601 760 176 064 × 2 = 0 + 0.003 520 352 035 203 520 352 128;
  • 6) 0.003 520 352 035 203 520 352 128 × 2 = 0 + 0.007 040 704 070 407 040 704 256;
  • 7) 0.007 040 704 070 407 040 704 256 × 2 = 0 + 0.014 081 408 140 814 081 408 512;
  • 8) 0.014 081 408 140 814 081 408 512 × 2 = 0 + 0.028 162 816 281 628 162 817 024;
  • 9) 0.028 162 816 281 628 162 817 024 × 2 = 0 + 0.056 325 632 563 256 325 634 048;
  • 10) 0.056 325 632 563 256 325 634 048 × 2 = 0 + 0.112 651 265 126 512 651 268 096;
  • 11) 0.112 651 265 126 512 651 268 096 × 2 = 0 + 0.225 302 530 253 025 302 536 192;
  • 12) 0.225 302 530 253 025 302 536 192 × 2 = 0 + 0.450 605 060 506 050 605 072 384;
  • 13) 0.450 605 060 506 050 605 072 384 × 2 = 0 + 0.901 210 121 012 101 210 144 768;
  • 14) 0.901 210 121 012 101 210 144 768 × 2 = 1 + 0.802 420 242 024 202 420 289 536;
  • 15) 0.802 420 242 024 202 420 289 536 × 2 = 1 + 0.604 840 484 048 404 840 579 072;
  • 16) 0.604 840 484 048 404 840 579 072 × 2 = 1 + 0.209 680 968 096 809 681 158 144;
  • 17) 0.209 680 968 096 809 681 158 144 × 2 = 0 + 0.419 361 936 193 619 362 316 288;
  • 18) 0.419 361 936 193 619 362 316 288 × 2 = 0 + 0.838 723 872 387 238 724 632 576;
  • 19) 0.838 723 872 387 238 724 632 576 × 2 = 1 + 0.677 447 744 774 477 449 265 152;
  • 20) 0.677 447 744 774 477 449 265 152 × 2 = 1 + 0.354 895 489 548 954 898 530 304;
  • 21) 0.354 895 489 548 954 898 530 304 × 2 = 0 + 0.709 790 979 097 909 797 060 608;
  • 22) 0.709 790 979 097 909 797 060 608 × 2 = 1 + 0.419 581 958 195 819 594 121 216;
  • 23) 0.419 581 958 195 819 594 121 216 × 2 = 0 + 0.839 163 916 391 639 188 242 432;
  • 24) 0.839 163 916 391 639 188 242 432 × 2 = 1 + 0.678 327 832 783 278 376 484 864;
  • 25) 0.678 327 832 783 278 376 484 864 × 2 = 1 + 0.356 655 665 566 556 752 969 728;
  • 26) 0.356 655 665 566 556 752 969 728 × 2 = 0 + 0.713 311 331 133 113 505 939 456;
  • 27) 0.713 311 331 133 113 505 939 456 × 2 = 1 + 0.426 622 662 266 227 011 878 912;
  • 28) 0.426 622 662 266 227 011 878 912 × 2 = 0 + 0.853 245 324 532 454 023 757 824;
  • 29) 0.853 245 324 532 454 023 757 824 × 2 = 1 + 0.706 490 649 064 908 047 515 648;
  • 30) 0.706 490 649 064 908 047 515 648 × 2 = 1 + 0.412 981 298 129 816 095 031 296;
  • 31) 0.412 981 298 129 816 095 031 296 × 2 = 0 + 0.825 962 596 259 632 190 062 592;
  • 32) 0.825 962 596 259 632 190 062 592 × 2 = 1 + 0.651 925 192 519 264 380 125 184;
  • 33) 0.651 925 192 519 264 380 125 184 × 2 = 1 + 0.303 850 385 038 528 760 250 368;
  • 34) 0.303 850 385 038 528 760 250 368 × 2 = 0 + 0.607 700 770 077 057 520 500 736;
  • 35) 0.607 700 770 077 057 520 500 736 × 2 = 1 + 0.215 401 540 154 115 041 001 472;
  • 36) 0.215 401 540 154 115 041 001 472 × 2 = 0 + 0.430 803 080 308 230 082 002 944;
  • 37) 0.430 803 080 308 230 082 002 944 × 2 = 0 + 0.861 606 160 616 460 164 005 888;

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:


0.000 110 011 001 100 110 011 004(10) =

0.0000 0000 0000 0111 0011 0101 1010 1101 1010 0(2)

6. Positive number before normalization:

0.000 110 011 001 100 110 011 004(10) =

0.0000 0000 0000 0111 0011 0101 1010 1101 1010 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 110 011 001 100 110 011 004(10) =

0.0000 0000 0000 0111 0011 0101 1010 1101 1010 0(2) =

0.0000 0000 0000 0111 0011 0101 1010 1101 1010 0(2) × 20 =

1.1100 1101 0110 1011 0110 100(2) × 2-14


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): -14

Mantissa (not normalized):
1.1100 1101 0110 1011 0110 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-14 + 2(8-1) - 1 =

(-14 + 127)(10) =

113(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;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

113(10) =

0111 0001(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 0110 1011 0101 1011 0100 =

110 0110 1011 0101 1011 0100


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) =
0111 0001

Mantissa (23 bits) =
110 0110 1011 0101 1011 0100


Decimal number -0.000 110 011 001 100 110 011 004 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0001 - 110 0110 1011 0101 1011 0100

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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