-0.000 000 100 016 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 100 016(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
-0.000 000 100 016(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:

|-0.000 000 100 016| = 0.000 000 100 016


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: 0.000 000 100 016.

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) 0.000 000 100 016 × 2 = 0 + 0.000 000 200 032;
  • 2) 0.000 000 200 032 × 2 = 0 + 0.000 000 400 064;
  • 3) 0.000 000 400 064 × 2 = 0 + 0.000 000 800 128;
  • 4) 0.000 000 800 128 × 2 = 0 + 0.000 001 600 256;
  • 5) 0.000 001 600 256 × 2 = 0 + 0.000 003 200 512;
  • 6) 0.000 003 200 512 × 2 = 0 + 0.000 006 401 024;
  • 7) 0.000 006 401 024 × 2 = 0 + 0.000 012 802 048;
  • 8) 0.000 012 802 048 × 2 = 0 + 0.000 025 604 096;
  • 9) 0.000 025 604 096 × 2 = 0 + 0.000 051 208 192;
  • 10) 0.000 051 208 192 × 2 = 0 + 0.000 102 416 384;
  • 11) 0.000 102 416 384 × 2 = 0 + 0.000 204 832 768;
  • 12) 0.000 204 832 768 × 2 = 0 + 0.000 409 665 536;
  • 13) 0.000 409 665 536 × 2 = 0 + 0.000 819 331 072;
  • 14) 0.000 819 331 072 × 2 = 0 + 0.001 638 662 144;
  • 15) 0.001 638 662 144 × 2 = 0 + 0.003 277 324 288;
  • 16) 0.003 277 324 288 × 2 = 0 + 0.006 554 648 576;
  • 17) 0.006 554 648 576 × 2 = 0 + 0.013 109 297 152;
  • 18) 0.013 109 297 152 × 2 = 0 + 0.026 218 594 304;
  • 19) 0.026 218 594 304 × 2 = 0 + 0.052 437 188 608;
  • 20) 0.052 437 188 608 × 2 = 0 + 0.104 874 377 216;
  • 21) 0.104 874 377 216 × 2 = 0 + 0.209 748 754 432;
  • 22) 0.209 748 754 432 × 2 = 0 + 0.419 497 508 864;
  • 23) 0.419 497 508 864 × 2 = 0 + 0.838 995 017 728;
  • 24) 0.838 995 017 728 × 2 = 1 + 0.677 990 035 456;
  • 25) 0.677 990 035 456 × 2 = 1 + 0.355 980 070 912;
  • 26) 0.355 980 070 912 × 2 = 0 + 0.711 960 141 824;
  • 27) 0.711 960 141 824 × 2 = 1 + 0.423 920 283 648;
  • 28) 0.423 920 283 648 × 2 = 0 + 0.847 840 567 296;
  • 29) 0.847 840 567 296 × 2 = 1 + 0.695 681 134 592;
  • 30) 0.695 681 134 592 × 2 = 1 + 0.391 362 269 184;
  • 31) 0.391 362 269 184 × 2 = 0 + 0.782 724 538 368;
  • 32) 0.782 724 538 368 × 2 = 1 + 0.565 449 076 736;
  • 33) 0.565 449 076 736 × 2 = 1 + 0.130 898 153 472;
  • 34) 0.130 898 153 472 × 2 = 0 + 0.261 796 306 944;
  • 35) 0.261 796 306 944 × 2 = 0 + 0.523 592 613 888;
  • 36) 0.523 592 613 888 × 2 = 1 + 0.047 185 227 776;
  • 37) 0.047 185 227 776 × 2 = 0 + 0.094 370 455 552;
  • 38) 0.094 370 455 552 × 2 = 0 + 0.188 740 911 104;
  • 39) 0.188 740 911 104 × 2 = 0 + 0.377 481 822 208;
  • 40) 0.377 481 822 208 × 2 = 0 + 0.754 963 644 416;
  • 41) 0.754 963 644 416 × 2 = 1 + 0.509 927 288 832;
  • 42) 0.509 927 288 832 × 2 = 1 + 0.019 854 577 664;
  • 43) 0.019 854 577 664 × 2 = 0 + 0.039 709 155 328;
  • 44) 0.039 709 155 328 × 2 = 0 + 0.079 418 310 656;
  • 45) 0.079 418 310 656 × 2 = 0 + 0.158 836 621 312;
  • 46) 0.158 836 621 312 × 2 = 0 + 0.317 673 242 624;
  • 47) 0.317 673 242 624 × 2 = 0 + 0.635 346 485 248;

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:


0.000 000 100 016(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1001 0000 1100 000(2)

6. Positive number before normalization:

0.000 000 100 016(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1001 0000 1100 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 100 016(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1001 0000 1100 000(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 1001 0000 1100 000(2) × 20 =

1.1010 1101 1001 0000 1100 000(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1010 1101 1001 0000 1100 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1100 1000 0110 0000 =

101 0110 1100 1000 0110 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 0111

Mantissa (23 bits) =
101 0110 1100 1000 0110 0000


Decimal number -0.000 000 100 016 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 101 0110 1100 1000 0110 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