-0.000 000 000 761 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 761(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
-0.000 000 000 761(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:

|-0.000 000 000 761| = 0.000 000 000 761


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: 0.000 000 000 761.

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) 0.000 000 000 761 × 2 = 0 + 0.000 000 001 522;
  • 2) 0.000 000 001 522 × 2 = 0 + 0.000 000 003 044;
  • 3) 0.000 000 003 044 × 2 = 0 + 0.000 000 006 088;
  • 4) 0.000 000 006 088 × 2 = 0 + 0.000 000 012 176;
  • 5) 0.000 000 012 176 × 2 = 0 + 0.000 000 024 352;
  • 6) 0.000 000 024 352 × 2 = 0 + 0.000 000 048 704;
  • 7) 0.000 000 048 704 × 2 = 0 + 0.000 000 097 408;
  • 8) 0.000 000 097 408 × 2 = 0 + 0.000 000 194 816;
  • 9) 0.000 000 194 816 × 2 = 0 + 0.000 000 389 632;
  • 10) 0.000 000 389 632 × 2 = 0 + 0.000 000 779 264;
  • 11) 0.000 000 779 264 × 2 = 0 + 0.000 001 558 528;
  • 12) 0.000 001 558 528 × 2 = 0 + 0.000 003 117 056;
  • 13) 0.000 003 117 056 × 2 = 0 + 0.000 006 234 112;
  • 14) 0.000 006 234 112 × 2 = 0 + 0.000 012 468 224;
  • 15) 0.000 012 468 224 × 2 = 0 + 0.000 024 936 448;
  • 16) 0.000 024 936 448 × 2 = 0 + 0.000 049 872 896;
  • 17) 0.000 049 872 896 × 2 = 0 + 0.000 099 745 792;
  • 18) 0.000 099 745 792 × 2 = 0 + 0.000 199 491 584;
  • 19) 0.000 199 491 584 × 2 = 0 + 0.000 398 983 168;
  • 20) 0.000 398 983 168 × 2 = 0 + 0.000 797 966 336;
  • 21) 0.000 797 966 336 × 2 = 0 + 0.001 595 932 672;
  • 22) 0.001 595 932 672 × 2 = 0 + 0.003 191 865 344;
  • 23) 0.003 191 865 344 × 2 = 0 + 0.006 383 730 688;
  • 24) 0.006 383 730 688 × 2 = 0 + 0.012 767 461 376;
  • 25) 0.012 767 461 376 × 2 = 0 + 0.025 534 922 752;
  • 26) 0.025 534 922 752 × 2 = 0 + 0.051 069 845 504;
  • 27) 0.051 069 845 504 × 2 = 0 + 0.102 139 691 008;
  • 28) 0.102 139 691 008 × 2 = 0 + 0.204 279 382 016;
  • 29) 0.204 279 382 016 × 2 = 0 + 0.408 558 764 032;
  • 30) 0.408 558 764 032 × 2 = 0 + 0.817 117 528 064;
  • 31) 0.817 117 528 064 × 2 = 1 + 0.634 235 056 128;
  • 32) 0.634 235 056 128 × 2 = 1 + 0.268 470 112 256;
  • 33) 0.268 470 112 256 × 2 = 0 + 0.536 940 224 512;
  • 34) 0.536 940 224 512 × 2 = 1 + 0.073 880 449 024;
  • 35) 0.073 880 449 024 × 2 = 0 + 0.147 760 898 048;
  • 36) 0.147 760 898 048 × 2 = 0 + 0.295 521 796 096;
  • 37) 0.295 521 796 096 × 2 = 0 + 0.591 043 592 192;
  • 38) 0.591 043 592 192 × 2 = 1 + 0.182 087 184 384;
  • 39) 0.182 087 184 384 × 2 = 0 + 0.364 174 368 768;
  • 40) 0.364 174 368 768 × 2 = 0 + 0.728 348 737 536;
  • 41) 0.728 348 737 536 × 2 = 1 + 0.456 697 475 072;
  • 42) 0.456 697 475 072 × 2 = 0 + 0.913 394 950 144;
  • 43) 0.913 394 950 144 × 2 = 1 + 0.826 789 900 288;
  • 44) 0.826 789 900 288 × 2 = 1 + 0.653 579 800 576;
  • 45) 0.653 579 800 576 × 2 = 1 + 0.307 159 601 152;
  • 46) 0.307 159 601 152 × 2 = 0 + 0.614 319 202 304;
  • 47) 0.614 319 202 304 × 2 = 1 + 0.228 638 404 608;
  • 48) 0.228 638 404 608 × 2 = 0 + 0.457 276 809 216;
  • 49) 0.457 276 809 216 × 2 = 0 + 0.914 553 618 432;
  • 50) 0.914 553 618 432 × 2 = 1 + 0.829 107 236 864;
  • 51) 0.829 107 236 864 × 2 = 1 + 0.658 214 473 728;
  • 52) 0.658 214 473 728 × 2 = 1 + 0.316 428 947 456;
  • 53) 0.316 428 947 456 × 2 = 0 + 0.632 857 894 912;
  • 54) 0.632 857 894 912 × 2 = 1 + 0.265 715 789 824;

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:


0.000 000 000 761(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0100 1011 1010 0111 01(2)

6. Positive number before normalization:

0.000 000 000 761(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0100 1011 1010 0111 01(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:


0.000 000 000 761(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0100 1011 1010 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0100 1011 1010 0111 01(2) × 20 =

1.1010 0010 0101 1101 0011 101(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):
1.1010 0010 0101 1101 0011 101


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. 101 0001 0010 1110 1001 1101 =

101 0001 0010 1110 1001 1101


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) =
101 0001 0010 1110 1001 1101


Decimal number -0.000 000 000 761 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0001 0010 1110 1001 1101

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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